2 Auxiliary results

According to the definition of $\mathrm{dom}(A_{0,1})$ , an important role is played in the forthcoming analysis by functions $f\in X_1$ such that $f''\in X_1$ . We collect useful properties of this class of functions in the next lemma and show, in particular, that the boundary condition (1.1c) is well defined for such functions.

Lemma 2.1 Consider $f\in X_1$ such that $f''\in X_1$ . Then $f\in BC([0,\infty))\cap C^1((0,\infty))$ , $f'\in L_1(0,\infty)$ , and, for $x> 0$ ,

(2.1a) \begin{equation} |f(x)| \le \|f''\|_{X_1}\,, \qquad x |f'(x)|\le \|f''\|_{X_1}\,, \qquad \|f'\|_{L_1(0,\infty)} \le \|f''\|_{X_1}\,. \end{equation}

Moreover,

(2.1b) \begin{equation} \lim_{x\to \infty} x f(x) = \lim_{x\to \infty} x f'(x) = 0\,. \end{equation}

In fact, f and $^f{\prime}$ are given by

(2.1c) \begin{equation} f(x) = - \int_x^\infty f'(y)\ \mathrm{d}y\,, \quad f'(x) = - \int_x^\infty f''(y)\ \mathrm{d}y\,, \qquad x\in (0,\infty)\,. \end{equation}

Also, $f\in X_m$ for any $m\in (-1,1)$ and, for all $\varepsilon>0$ ,

(2.2) \begin{equation} \|f\|_{X_m} \le \frac{\varepsilon^{m+1}}{m+1} \|f''\|_{X_1} + \varepsilon^{m-1} \|f\|_{X_1}\,. \end{equation}

Equivalently,

(2.3) \begin{equation} \|f\|_{X_m} \le \frac{2 (1-m)^{(m-1)/2}}{m+1} \|f''\|_{X_1}^{(1-m)/2} \|f\|_{X_1}^{(m+1)/2}\,,\qquad m\in (-1,1)\,. \end{equation}

Proof. Introducing

\begin{equation*} F(x) := \int_x^\infty (y-x) f''(y)\ \mathrm{d}y\,, \qquad x\in (0,\infty)\,, \end{equation*}

it follows from the integrability of f $^{\prime\prime}$ that

\begin{equation*} - \|f''\|_{X_1} \le - \int_x^\infty (y-x) |f''(y)| \mathrm{d}y \le F(x) \le \int_x^\infty (y-x) |f''(y)| \mathrm{d}y \le \|f''\|_{X_1}\,, \end{equation*}

so that F(x) is well defined for $x\ge 0$ . Moreover,

\begin{equation*} F\in BC([0,\infty))\cap C^1((0,\infty))\cap W_{1,loc}^2(0,\infty) \end{equation*}

and satisfies

(2.4) \begin{equation} F'(x) = - \int_x^\infty f''(y)\ \mathrm{d}y\,, \qquad F''(x) = f''(x)\,, \qquad x\in (0,\infty)\,, \end{equation}

and

(2.5) \begin{equation} \lim_{x\to \infty} x F'(x) = \lim_{x\to \infty} F(x) = 0\,. \end{equation}

In particular, we infer from (2.4) that there is $(\alpha,\beta)\in\mathbb{R}^2$ such that $(f-F)(x) = \alpha + \beta x$ for $x>0$ . Moreover, since $f\in X_1$ , it follows from (2.5) that

\begin{equation*} \lim_{x\to\infty} \int_x^{x+1} |\alpha + \beta y|\ \mathrm{d}y \le \lim_{x\to\infty} \int_x^{x+1} (|f(y)|+|F(y)|)\ \mathrm{d}y = 0\,, \end{equation*}

which readily gives $\alpha=\beta=0$ and $F=f$ , thereby establishing (2.1), except for the limiting behavior of f at infinity. To this end, we observe that, since $f\in L_\infty(0,\infty)$ and $f'\in L_1(0,\infty)$ , a formula for $f(x)^2$ reads

\begin{equation*} f(x)^2 = - 2 \int_x^\infty f(y) f'(y)\ \mathrm{d}y\,, \qquad x\ge 0\,. \end{equation*}

We then deduce from (2.1a) that

\begin{equation*} f(x)^2 \le 2 \int_x^\infty y |f'(y)| |f(y)| \ \frac{\mathrm{d}y}{y} \le \frac{2}{x^2} \|f''\|_{X_1} \int_x^\infty y |f(y)|\ \mathrm{d}y\,, \end{equation*}

which implies that $x f(x)\to 0$ as $x\to\infty$ due to $f\in X_1$ .

Next, let $\varepsilon>\delta>0$ . It follows from (2.1c) that

\begin{align*} \int_\delta^\infty x^m |f(x)|\ \mathrm{d}x & \le \int_\delta^\varepsilon x^m |f(x)|\ \mathrm{d}x + \varepsilon^{m-1} \int_\varepsilon^\infty x\ |f(x)|\ \mathrm{d}x \\[3pt] & \le \int_\delta^\varepsilon x^m \int_x^\infty |f'(y)|\ \mathrm{d}y \mathrm{d}x + \varepsilon^{m-1} \|f\|_{X_1} \\[3pt] & \le \int_\delta^\varepsilon x^m \int_x^\varepsilon |f'(y)|\ \mathrm{d}y \mathrm{d}x + \frac{\varepsilon^{m+1}}{m+1} \int_\varepsilon^\infty |f'(y)|\ \mathrm{d}y + \varepsilon^{m-1} \|f\|_{X_1} \,. \end{align*}

By Fubini’s theorem,

\begin{equation*} \int_\delta^\varepsilon x^m \int_x^\varepsilon |f'(y)|\ \mathrm{d}y \mathrm{d}x = \frac{1}{m+1} \int_\delta^\varepsilon \left( y^{m+1} - \delta^{m+1} \right) |f'(y)|\ \mathrm{d}y \le \frac{\varepsilon^{m+1}}{m+1} \int_0^\varepsilon |f'(y)|\ \mathrm{d}y \,. \end{equation*}

Combining the above inequalities with (2.1a) gives

\begin{equation*} \int_\delta^\infty x^m |f(x)|\ \mathrm{d}x \le \frac{\varepsilon^{m+1}}{m+1} \int_0^\infty |f'(y)|\ \mathrm{d}y + \varepsilon^{m-1} \|f\|_{X_1} \le \frac{\varepsilon^{m+1}}{m+1} \|f''\|_{X_1} + \varepsilon^{m-1} \|f\|_{X_1}\,. \end{equation*}

Letting $\delta\to 0$ completes the proof of (2.2). We next optimize (2.2) with respect to $\varepsilon\in (0,\infty)$ to derive (2.3).

We next state for the sake of completeness the density in $X_{1,m}$ of smooth functions with compact support in $(0,\infty)$ , which is actually a straightforward consequence of the density of $C_c^\infty((0,\infty))$ in $L_1(0,\infty)$ .

We finally recall a variant of the celebrated inequality of Kato [Reference Kato23, Lemma A].

Lemma 2.3 Let $\ell$ be a nonnegative function in $W_{\infty,loc}^1([0,\infty))$ with $\ell(0)=0$ and consider $f\in W_{1,loc}^2(0,\infty) \cap X_1$ such that $f'\ell'\in L_1((0,\infty))$ and $f''\in L_1((0,\infty),\ell(x)\mathrm{d}x)$ . Then

(2.6) \begin{equation} - \int_0^\infty \ell(x)\ \mathrm{sign}(f(x)) f''(x)\ \mathrm{d}x \ge \int_0^\infty \ell'(x)\ |f|'(x)\ \mathrm{d}x\,. \end{equation}

Proof. For $\varepsilon\in (0,1)$ , we define $\beta_\varepsilon\in W_{\infty,loc}^2(\mathbb{R})$ by $\beta_\varepsilon(0)=0$ and

\begin{equation*} \beta_\varepsilon^{\prime}(r) := \mathrm{sign}(r)\,, \quad |r|>\varepsilon\,, \;\;\text{ and }\;\; \beta_\varepsilon^{\prime}(r) := \frac{r}{\varepsilon}\,, \quad r\in [-\varepsilon,\varepsilon]\,. \end{equation*}

Since $f\in X_1$ , we note that

\begin{equation*} |\beta_\varepsilon^{\prime}(f)| \le \frac{|f|}{\varepsilon} \in X_1\,, \end{equation*}

while the integrability assumptions on f and $\ell$ imply that

\begin{align*} |(\ell f')(x)| & \le |(\ell f')(1)| + \int_1^x |(\ell' f' + \ell f'')(y)|\ \mathrm{d}y \\[3pt] & \le |(\ell f')(1)| + \|\ell' f'\|_{L_1(0,\infty)} + \|\ell f''\|_{L_1(0,\infty)}\,, \qquad x\ge 1\,. \end{align*}

Next, integration by parts gives

\begin{equation*} - \int_0^\infty \ell(x)\ \beta_\varepsilon^{\prime}(f(x)) f''(x)\ \mathrm{d}x = \int_0^\infty \left[ \ell(x)\ \beta_\varepsilon^{\prime\prime}(f(x)) |f'(x)|^2 + \ell'(x) \beta_\varepsilon^{\prime}(f(x)) f'(x) \right]\ \mathrm{d}x \,, \end{equation*}

observing that the boundary terms vanish due to $\ell(0)=0$ , $\ell f'\in L_\infty(1,\infty)$ , and $\beta_\varepsilon^{\prime}(f)\in L_1(1,\infty)$ . We then deduce from the nonnegativity of $\beta_\varepsilon^{\prime\prime}$ on $\mathbb{R}$ that

\begin{equation*} - \int_0^\infty \ell(x)\ \beta_\varepsilon^{\prime}(f(x)) f''(x)\ \mathrm{d}x \ge \int_0^\infty \ell'(x) \beta_\varepsilon^{\prime}(f(x)) f'(x)\ \mathrm{d}x\,. \end{equation*}

Since $|\beta_\varepsilon(r)-|r||\le \varepsilon$ for $r\in\mathbb{R}$ and since $\beta_\varepsilon^{\prime}$ converges pointwise to the sign function in $\mathbb{R}$ as $\varepsilon \to 0$ with $|\beta_\varepsilon^{\prime}|\le 1$ , we may pass to the limit as $\varepsilon \to 0$ in the previous inequality with the help of Lebesgue’s dominated convergence theorem and the integrability properties of f $^{\prime}$ and f $^{\prime\prime}$ and find

\begin{equation*} - \int_0^\infty \ell(x)\ \mathrm{sign}(f(x)) f''(x)\ \mathrm{d}x \ge \int_0^\infty \ell'(x) \mathrm{sign}(f(x)) f'(x)\ \mathrm{d}x\,. \end{equation*}

We finally use the classical property $|f|'=\mathrm{sign}(f) f'$ a.e. in $(0,\infty)$ , see [Reference Kinderlehrer and Stampacchia24, Chapter II, Theorem A.2], to complete the proof.

3 The heat semigroup

It is well known, see [Reference Craig13, Section 3.4] for instance, that the solution to the heat equation with homogeneous Dirichlet boundary conditions at $x=0$ ,

\begin{align*} \partial_t w - \partial_x^2 w & = 0\,, \qquad (t,x)\in (0,\infty)\times (0,\infty)\,, \\[3pt] w(t,0) & = 0\,, \qquad t\in (0,\infty)\,, \\[3pt] w(0,x) & = f(x)\,, \qquad x\in (0,\infty)\,,\end{align*}

is given by the representation formula

(3.1) \begin{equation} w(t,x) = W(t)f(x) := \int_0^\infty [k(t,x-y)-k(t,x+y)] f(y)\ \mathrm{d}y\,, \quad (t,x)\in (0,\infty)^2\,, \end{equation}

where

(3.2) \begin{equation} k(t,x):=\frac{1}{\sqrt{4\pi t}} e^{-\vert x\vert^2/4t}\,,\qquad (t,x)\in (0,\infty)\times \mathbb{R}\,.\end{equation}

Moreover, $(W(t))_{t\ge 0}$ (with $W(0):=\mathrm{id}_{L_1(0,\infty)}$ ) is a positive analytic semigroup of contractions on $L_1(0,\infty)$ with generator G given by

\begin{align*} \mathrm{dom}(G) & := \{ f\in L_1(0,\infty)\ : \ f''\in L_1(0,\infty)\;\text{ and }\; f(0) = 0 \}\,, \\ G f & := f''\,, \qquad f\in \mathrm{dom}(G)\,. \end{align*}

That is, $G\in\mathcal{G}_+(L_1(0,\infty),1,0)\cap \mathcal{H}(L_1(0,\infty))$ and $e^{tG}=W(t)$ for $t\ge 0$ . We now consider this semigroup in the weighted space $X_{1,m}$ . More precisely, for $m\ge 1$ , we recall the definition (1.4) (with $a=0$ ) of the unbounded operator $A_{0,m}$ on $X_{1,m}$ , given by

(3.3) \begin{align} \mathrm{dom}(A_{0,m}) & = \{ f\in X_{1,m}\ : \ f''\in X_{1,m}\;\text{ and }\; f(0) = 0 \}\,, \nonumber\\[3pt] A_{0,m} f & = f''\,, \qquad f\in \mathrm{dom}(A_{0,m})\,, \end{align}

and show that it is the generator of the heat semigroup in $X_{1,m}$ .

Proposition 3.1 Let $m\ge 1$ . There is $\omega_m\ge 0$ such $A_{0,m}\in \mathcal{G}_+(X_{1,m},1,\omega_m)\cap \mathcal{H}(X_{1,m})$ with $(0,\infty)\subset \rho(A_{0,m})$ . The semigroup $\left( e^{tA_{0,m}} \right)_{t\ge 0}$ is given by

(3.4) \begin{equation} e^{tA_{0,m}}f (x) = \int_0^\infty [k(t,x-y)-k(t,x+y)] f(y)\ \mathrm{d}y\,, \qquad (t,x)\in (0,\infty)^2\,, \end{equation}

for $f\in X_{1,m}$ , where k is defined in (3.2). Moreover, $e^{tA_{0,m}} = e^{tA_{0,1}}|_{X_{1,m}}$ for $t\ge 0$ and

\begin{equation*} \left( \lambda - A_{0,m} \right)^{-1} = \left( \lambda - A_{0,1} \right)^{-1}|_{X_{1,m}}\,, \qquad \lambda>0\,. \end{equation*}

Two steps are needed to show Proposition 3.1. We first establish Proposition 3.1 for $m=1$ and $m\ge 3$ , see Lemma 3.2 below. An interpolation argument then completes the proof for $m\in (1,3)$ .

Lemma 3.2 Let $m\in \{1\} \cup (3,\infty)$ . Then $A_{0,m}\in \mathcal{G}_+(X_{1,m},1,\omega_m)\cap \mathcal{H}(X_{1,m})$ with

\begin{equation*} \omega_1 := 0 \;\text{ and }\; \omega_m := 4^{1/(m-1)} m (m-3)^{(m-3)/(m-1)}\,, \qquad m\ge 3\,. \end{equation*}

Moreover, $e^{tA_{0,m}} = e^{tA_{0,1}}|_{X_{1,m}}$ for $t\ge 0$ , $(0,\infty)\subset \rho(A_{0,m})$ , and

\begin{equation*} (\lambda - A_{0,m})^{-1} = (\lambda-A_{0,1})^{-1}|_{X_{1,m}}\,,\qquad \lambda>0\,. \end{equation*}

Proof. We first note that $A_{0,m}$ is a closed operator on $X_{1,m}$ and that its domain is dense in $X_{1,m}$ due to Lemmas 2.1 and 2.2. We divide the remainder of the proof into several steps.

Step 1. We first show the dissipativity of $A_{0,m}-\omega_m$ on $X_{1,m}$ . To this end, let $\lambda>0$ and $f\in \mathrm{dom}(A_{0,m})$ . Using the inequality

\begin{equation*} |r-s| \ge \mathrm{sign}(r) (r-s)\,, \qquad (r,s)\in \mathbb{R}^2\,, \end{equation*}

along with the nonnegativity of $\lambda$ and $\omega_m$ , gives

\begin{align*} \|\lambda f - (A_{0,m}-\omega_m)f\|_{X_m} & = \|(\lambda+\omega_m) f - A_{0,m} f\|_{X_m}\\ & \ge \int_0^\infty x^m\ \mathrm{sign}(f(x)) [(\lambda+\omega_m) f(x)-f''(x)]\ \mathrm{d}x\,. \end{align*}

By Kato’s inequality (2.6) (with $\ell(x) = x^m$ ), Lemma 2.1, and the boundary condition $f(0)=0$ , we further obtain

\begin{align*} \|\lambda f - (A_{0,m}-\omega_m)f\|_{X_m} & \ge (\lambda+\omega_m) \|f\|_{X_m} + m \int_0^\infty x^{m-1}\ |f|'(x)\ \mathrm{d}x \\ & = (\lambda+\omega_m) \|f\|_{X_m} - m(m-1) \|f\|_{X_{m-2}}\,. \end{align*}

In particular, when $m=1$ ,

(3.5) \begin{equation} \| \lambda f - A_{0,1}\,f\|_{X_1} \ge \lambda \|f\|_{X_1}\,, \end{equation}

so that $A_{0,1}$ is dissipative on $X_1$ . We next handle the case $m\ge 3$ . Then $m-2\in [1,m)$ and we infer from Young’s inequality that, for $\varepsilon>0$ ,

\begin{equation*} m(m-1) \|f\|_{X_{m-2}} \le m (m-3) \varepsilon \|f\|_{X_m} + 2m \varepsilon^{(3-m)/2} \|f\|_{X_1}\,. \end{equation*}

Hence,

\begin{equation*} (\lambda+\omega_m) \|f\|_{X_m} \le m (m-3) \varepsilon \|f\|_{X_m} + 2m \varepsilon^{(3-m)/2} \|f\|_{X_1} + \|\lambda f - (A_{0,m}-\omega_m)f\|_{X_m}\,. \end{equation*}

Combining (3.5) with this inequality gives

\begin{align*} (\lambda+\omega_m) \|f\|_{X_{1,m}} & \le \|\lambda f - (A_{0,m}-\omega_m) f\|_{X_{1,m}} \\ & \qquad + m (m-3) \varepsilon \|f\|_{X_m} + 2m \varepsilon^{(3-m)/2} \|f\|_{X_1}\,. \end{align*}

We now choose $\varepsilon = \varepsilon_m := (2/(m-3))^{2/(m-1)}$ . Since

\begin{equation*} \omega_m = m (m-3) \varepsilon_m = m \varepsilon_m^{(3-m)/2}\,, \end{equation*}

we readily conclude that

\begin{equation*} \lambda \|f\|_{X_{1,m}} \le \|\lambda f - (A_{0,m}-\omega_m ) f\|_{X_{1,m}} \,, \end{equation*}

so that $A_{0,m}-\omega_m I$ is a dissipative operator on $X_{1,m}$ .

Step 2. We next show that $\mathrm{rg}(\lambda - A_{0,m})=X_{1,m}$ for $\lambda>0$ . Consider $g\in X_{1,m}$ . According to Lemma 2.2, there is a sequence $(g_n)_{n\ge 1}$ in $C_c^\infty((0,\infty))$ such that

(3.6) \begin{equation} \lim_{n\to\infty} \|g_n-g\|_{X_{1,m}} = 0\,. \end{equation}

Since $g_n\in L_1(0,\infty)$ and $(0,\infty)\subset \rho(G)$ , there is a unique $f_n\in \mathrm{dom}(G)$ such that $\lambda f_n - G f_n = g_n$ ; that is,

(3.7) \begin{equation} \lambda f_n - f_n^{\prime\prime} = g_n \;\;\text{ in }\;\; (0,\infty)\,, \qquad f_n(0)=0\,. \end{equation}

Now, let $R>1$ . We multiply (3.7) by $(x\wedge R)^m\ \mathrm{sign}(f_n(x))$ and integrate over $(0,\infty)$ . Using Kato’s inequality (2.6) (with $\ell(x)= (x\wedge R)^m$ ), we obtain

\begin{align*} \|g_n\|_{X_m} & \ge \int_0^\infty (x\wedge R)^m\ \mathrm{sign}(f_n(x)) [\lambda f_n(x)-f_n^{\prime\prime}(x)]\ \mathrm{d}x \\[3pt] & \ge \lambda \int_0^\infty (x\wedge R)^m\ |f_n(x)|\ \mathrm{d}x + m \int_0^R x^{m-1} |f_n|'(x)\ \mathrm{d}x \\[3pt] & = \lambda \int_0^\infty (x\wedge R)^m\ |f_n(x)|\ \mathrm{d}x - m(m-1) \int_0^R x^{m-2}\ f_n(x)\ \mathrm{d}x \,. \end{align*}

In particular, for $m=1$ we get

\begin{equation*} \|g_n\|_{X_1} \ge \lambda \int_0^\infty (x\wedge R)\ |f_n(x)|\ \mathrm{d}x \,, \end{equation*}

so that, letting $R\to \infty$ and using Fatou’s lemma,

(3.8) \begin{equation} \lambda \|f_n\|_{X_1} \le \|g_n\|_{X_1}\,. \end{equation}

The same argument entails that

(3.9) \begin{equation} \lambda \|f_n-f_l\|_{X_1} \le \|g_n-g_l\|_{X_1}\,, \qquad n, l\ge 1\,. \end{equation}

If $m\ge 3$ , then $m-2\in [1,m)$ , and we use Young’s inequality to deduce that, for $\varepsilon>0$ ,

\begin{align*} \lambda \int_0^\infty (x\wedge R)^m\ |f_n(x)|\ \mathrm{d}x & \le \|g_n\|_{X_m} + m(m-3)\varepsilon \int_0^R x^m\ |f_n(x)|\ \mathrm{d}x\\ & \qquad + 2m \varepsilon^{(3-m)/2} \int_0^R x\ |f_n(x)|\ \mathrm{d}x \\ & \le \|g_n\|_{X_m} + m(m-3)\varepsilon \int_0^\infty (x\wedge R)^m\ |f_n(x)|\ \mathrm{d}x\\ & \qquad + 2m \varepsilon^{(3-m)/2} \|f_n\|_{X_1}\,. \end{align*}

Choosing $\varepsilon = \lambda/(2m(m-3))$ , we combine (3.8) and the above inequality to conclude that

\begin{equation*} \frac{\lambda}{2} \int_0^\infty (x\wedge R)^m\ |f_n(x)|\ \mathrm{d}x \le \|g_n\|_{X_m} + \frac{2m}{\lambda} \left( \frac{2m(m-3)}{\lambda} \right)^{(m-3)/2} \|g_n\|_{X_1}\,. \end{equation*}

We now let $R\to\infty$ in this inequality and deduce from Fatou’s lemma that $f_n\in X_m$ with

(3.10) \begin{equation} \lambda \|f_n\|_{X_m} \le 2 \|g_n\|_{X_m} + \frac{4m}{\lambda} \left( \frac{2m(m-3)}{\lambda} \right)^{(m-3)/2} \|g_n\|_{X_1}\,. \end{equation}

The same argument entails that

(3.11) \begin{equation} \lambda \|f_n-f_l\|_{X_m} \le 2 \|g_n-g_l\|_{X_m} + \frac{4m}{\lambda} \left( \frac{2m(m-3)}{\lambda} \right)^{(m-3)/2} \|g_n-g_l\|_{X_1} \end{equation}

for $n\ge 1$ and $l\ge 1$ . Therefore, for $m\in \{1\}\cup [3,\infty)$ , the estimates (3.9) and (3.11) along with (3.6) guarantee that $(f_n)_{n\ge 1}$ is a Cauchy sequence in $X_{1,m}$ and that there is $f\in X_{1,m}$ such that

(3.12) \begin{equation} \lim_{n\to\infty} \|f_n-f\|_{X_{1,m}} = 0\,. \end{equation}

Since $f_n^{\prime\prime}=\lambda f_n-g_n$ for all $n\ge 1$ by (3.7), it readily follows from (3.6) and (3.12) that $(f_n^{\prime\prime})_{n\ge 1}$ converges to $\lambda f-g$ in $X_{1,m}$ and to f $^{\prime\prime}$ in the sense of distributions. Therefore, $f''\in X_{1,m}$ , $f''=\lambda f-g$ , and $\|f_n^{\prime\prime}-f''\|_{X_{1,m}}\to 0$ as $n\to\infty$ . Finally, by (2.1c),

\begin{align*} |f(0)| & = |f(0)-f_n(0)| \le \int_0^\infty |(f'-f_n^{\prime})(x)|\ \mathrm{d}x \\ & \le \int_0^\infty \int_x^\infty |(f''-f_n^{\prime\prime})(y)|\ \mathrm{d}y\mathrm{d}x = \int_0^\infty y\ |(f''-f_n^{\prime\prime})(y)|\ \mathrm{d}y = \|f''-f_n^{\prime\prime}\|_{X_1}\,, \end{align*}

from which we deduce that $f(0)=0$ . Consequently, $f\in \mathrm{dom}(A_{0,m})$ and $\lambda f - A_{0,m}\, f = g$ . Since $\lambda - A_{0,m}$ is one-to-one by (3.5), we have thus shown for any $\lambda>0$ that

(3.13) \begin{equation} \mathrm{rg}(\lambda - A_{0,m})=X_{1,m} \;\text{ and }\; (\lambda-A_{0,m})^{-1}g = (\lambda - A_{0,1})^{-1}g \;\text{ for all }\; g\in X_{1,m}\,. \end{equation}

Step 3. For $m\in \{1\}\cup [3,\infty)$ , we infer from Step 1 to Step 2 and the Lumer-Phillips theorem [Reference Pazy30, Theorem 1.4.3] that $A_{0,m}-\omega_m$ belongs to $\mathcal{G}(X_{1,m},1,0)$ . Hence, $A_{0,m}\in \mathcal{G}(X_{1,m},1,\omega_m)$ . Also, it readily follows from (3.13) that $(0,\infty)\subset \rho(A_{0,m})$ and

\begin{equation*} (\lambda - A_{0,m})^{-1} = (\lambda-A_{0,1})^{-1}|_{X_{1,m}}\,. \end{equation*}

In particular, the latter, along with the exponential formula [Reference Pazy30, Theorem 1.8.3], entails that $e^{tA_{0,m}} = e^{tA_{0,1}}|_{X_{1,m}}$ for all $t\ge 0$ .

Step 4. We next derive the representation formulation for $(e^{tA_{0,1}})_{t\ge 0}$ . To this end, we first note that, for $t>0$ , the operator W(t) defined in (1.3) is a bounded operator on $X_1$ . Indeed, since $|x-y|\le |x+y|=x+y$ for $(x,y)\in (0,\infty)^2$ ,

(3.14) \begin{equation} k(t,x-y) - k(t,x+y) =\frac{1}{\sqrt{4\pi t}} \left( e^{-|x-y|^2/4t} - e^{-|x+y|^2/4t} \right)\ge 0 \end{equation}

for $ (x,y)\in (0,\infty)^2$ . By Fubini–Tonelli’s theorem and (3.14),

\begin{align*} \|W(t)f\|_{X_1} & \le \int_0^\infty x \int_0^\infty [k(t,x-y) - k(t,x+y)] |f(y)|\ \mathrm{d}y\mathrm{d}x \\[3pt] & = \frac{1}{\sqrt{\pi}} \int_0^\infty |f(y)| \int_{-y/2\sqrt{t}}^\infty (y+2z\sqrt{t}) e^{-z^2}\ \mathrm{d}z \mathrm{d}y \\[3pt] & \qquad - \frac{1}{\sqrt{\pi}} \int_0^\infty |f(y)| \int_{-\infty}^{-y/2\sqrt{t}} (-y-2z\sqrt{t}) e^{-z^2}\ \mathrm{d}z \mathrm{d}y \\[3pt] & = \frac{1}{\sqrt{\pi}} \int_0^\infty |f(y)| \left( \int_{-\infty}^\infty (y+2z\sqrt{t}) e^{-z^2}\ \mathrm{d}z \right)\ \mathrm{d}y = \|f\|_{X_1}\,. \end{align*}

Thus, W(t) is a contraction on $X_1$ . Since $(\lambda - G)^{-1} = (\lambda - A_{0,1})^{-1}$ on $L_1(0,\infty)\cap X_1$ for all $\lambda>0$ , it follows from the exponential formula, see [Reference Pazy30, Theorem 1.8.3], that $e^{tG}=e^{tA_{0,1}}$ on $L_1(0,\infty)\cap X_1$ for all $t\ge 0$ . Therefore, $W(t)=e^{tA_{0,1}}$ on $L_1(0,\infty)\cap X_1$ for all $t\ge 0$ . Since $L_1(0,\infty)\cap X_1$ is dense in $X_1$ by Lemma 2.2 and $e^{tA_{0,1}}$ and W(t) are both bounded operators on $X_1$ , we conclude that $e^{tA_{0,1}}=W(t)$ on $X_1$ for all $t\ge 0$ . Together with the outcome of Step 3, this identity proves (3.4). In particular, $(e^{tA_{0,m}})_{t\ge 0}$ is a positive semigroup according to (3.14).

Step 5. We are left with showing the analyticity of $(e^{tA_{0,m}})_{t\ge 0}$ on $X_{1,m}$ . To this end, let $f\in X_{1,m}$ . Clearly,

\begin{equation*} t \longmapsto \int_0^\infty [k(t,x-y)-k(t,x+y)] f(y)\ \mathrm{d}y \end{equation*}

is differentiable on $(0,\infty)$ and, since

\begin{equation*} 2t \partial_t k(t,x) = \left( - 1 + \frac{|x|^2}{2t} \right) k(t,x)\,, \qquad (t,x)\in (0,\infty)\times\mathbb{R}\,, \end{equation*}

we derive for $(t,x)\in (0,\infty)^2$ that

\begin{align*} 2t\frac{\mathrm{d}}{\mathrm{d}t} e^{tA_{0,m}}f(x) & = - e^{tA_{0,m}}f(x) \\[3pt] & \qquad + 2 \int_0^\infty \left( \frac{|x-y|^2}{4t} k(t,x-y) - \frac{|x+y|^2}{4t} k(t,x+y) \right) f(y)\ \mathrm{d}y \\[3pt] & = - 3 e^{tA_{0,m}}f(x) \\[3pt] & \qquad + 2 \int_0^\infty \left( 1 + \frac{|x-y|^2}{4t} \right) k(t,x-y) f(y)\ \mathrm{d}y \\[3pt] & \qquad - 2 \int_0^\infty \left( 1+ \frac{|x+y|^2}{4t} \right) k(t,x+y) f(y)\ \mathrm{d}y\,. \end{align*}

Let $t>0$ . Since $z\mapsto (1+z) e^{-z}$ is non-increasing on $(0,\infty)$ and $|x-y|\le x+y$ for $(x,y)\in (0,\infty)^2$ , we see that

(3.15) \begin{equation} \left( 1 + \frac{|x-y|^2}{4t} \right) k(t,x-y) - \left( 1+ \frac{|x+y|^2}{4t} \right) k(t,x+y) \ge 0 \end{equation}

for $(x,y)\in (0,\infty)^2$ and infer from Fubini–Tonelli’s theorem that

\begin{align*} & \int_0^\infty x^m \left| \int_0^\infty \left[ \left( 1 + \frac{|x-y|^2}{4t} \right) k(t,x-y) - \left( 1+ \frac{|x+y|^2}{4t} \right) k(t,x+y) \right] f(y)\ \mathrm{d}y \right|\ \mathrm{d}x \\[6pt] & \le \int_0^\infty x^m \int_0^\infty \left[ \left( 1 + \frac{|x-y|^2}{4t} \right) k(t,x-y) - \left( 1+ \frac{|x+y|^2}{4t} \right) k(t,x+y) \right] |f(y)|\ \mathrm{d}y\mathrm{d}x \\[6pt] & = \frac{1}{\sqrt{\pi}} \int_0^\infty |f(y)| \left[ \int_{-y/2\sqrt{t}}^\infty \left( y+2z\sqrt{t} \right)^m (1+z^2) e^{-z^2}\ \mathrm{d}z \right. \\[6pt] &\qquad\qquad\qquad\quad\qquad\qquad\quad \left. - \int_{-\infty}^{-y/2\sqrt{t}} \left( -y-2z\sqrt{t} \right)^m (1+z^2) e^{-z^2}\ \mathrm{d}z \right]\ \mathrm{d}y \\[6pt] & = \frac{1}{\sqrt{\pi}} \int_0^\infty |f(y)| \int_{-\infty}^\infty \left| y+2z\sqrt{t} \right|^{m-1} \left( y+2z\sqrt{t} \right) (1+z^2) e^{-z^2}\ \mathrm{d}z\mathrm{d}y \,. \end{align*}

Consequently,

\begin{align*} & 2t \left\| \frac{\mathrm{d}}{\mathrm{d}t} e^{tA_{0,m}}f \right\|_{X_{1,m}} \\[3pt] & \qquad \le 3 \left\| e^{tA_{0,m}}f \right\|_{X_{1,m}} + \frac{2}{\sqrt{\pi}} \int_0^\infty |f(y)| \int_{-\infty}^\infty \left( y+2z\sqrt{t} \right) (1+z^2) e^{-z^2}\ \mathrm{d}z\mathrm{d}y \\ & \qquad\qquad + \frac{2}{\sqrt{\pi}} \int_0^\infty |f(y)| \int_{-\infty}^\infty \left| y+2z\sqrt{t} \right|^{m-1} \left( y+2z\sqrt{t} \right) (1+z^2) e^{-z^2}\ \mathrm{d}z \, \mathrm{d}y\,, \end{align*}

so that

\begin{equation*} \limsup_{t\to 0} t \left\| \frac{\mathrm{d}}{\mathrm{d}t} e^{tA_{0,m}}f \right\|_{X_{1,m}} \le 3 \|f\|_{X_{1,m}} \end{equation*}

by Lebesgue’s convergence theorem. It then follows from [Reference Engel and Nagel14, Theorem II.4.6 (c)] that this property implies the analyticity of $(e^{tA_{0,m}})_{t\ge 0}$ on $X_{1,m}$ . Thus, the proof is complete.

The proof of Proposition 3.1 is now a consequence of the previous lemma and an interpolation argument as shown next.

Proof of Proposition 3.1. We only have to consider the case $m\in (1,3)$ . Since $A_{0,1}\in\mathcal{H}( X_{1,1})$ and $A_{0,3}\in\mathcal{H}(X_{1,3})$ according to Lemma 3.2, Hille’s characterization implies that there are $\lambda_0>0$ and $\kappa\ge 1$ such that

\begin{equation*} \|(\lambda-A_{0,1})^{-1}\|_{\mathcal{L}(X_{1,1})} +\|(\lambda-A_{0,3})^{-1}\|_{\mathcal{L}(X_{1,3})}\le \frac{\kappa}{\vert\lambda-\lambda_0\vert}\,,\quad \textsf{Re}\, \lambda> \lambda_0\,, \end{equation*}

see [Reference Engel and Nagel14, Theorem II.4.6 (d)] for instance. Since $(\lambda-A_{0,3})^{-1}=(\lambda-A_{0,1})^{-1}|_{X_{1,3}}$ , it readily follows by interpolation that

(3.16) \begin{equation} \|(\lambda-A_{0,1})^{-1}|_{X_{1,m}}\|_{\mathcal{L}(X_{1,m})} \le \frac{\kappa}{\vert\lambda-\lambda_0\vert}\,,\quad \textsf{Re}\, \lambda> \lambda_0\,. \end{equation}

Obviously, $A_{0,m}$ is closed and densely defined in $X_{1,m}$ . Let $g\in X_{1,m}$ be arbitrary and $\textsf{Re}\, \lambda> \lambda_0$ . There is a unique $f\in \mathrm{dom}(A_{0,1})$ such that $(\lambda-A_{0,1})f=g$ . By (3.16), $f=(\lambda-A_{0,1})^{-1}|_{X_{1,m}}g\in X_{1,m}$ and $f''=A_{0,m}f=\lambda f-g\in X_{1,m}$ . From this identity, we deduce $f\in \mathrm{dom}(A_{0,m})$ with $A_{0,m} f=A_{0,1}f$ and $(\lambda-A_{0,m})f=g$ . Since f is unique, we conclude that $\lambda\in\rho(A_{0,m})$ and

\begin{equation*} (\lambda-A_{0,m})^{-1}g=f=(\lambda-A_{0,1})^{-1}|_{X_{1,m}}g\,; \end{equation*}

that is, $(\lambda-A_{0,m})^{-1}=(\lambda-A_{0,1})^{-1}|_{X_{1,m}}$ . Invoking (3.16) we derive that

\begin{equation*} \|(\lambda-A_{0,m})^{-1}\|_{\mathcal{L}(X_{1,m})} \le \frac{\kappa}{\vert\lambda-\lambda_0\vert}\,,\quad \textsf{Re}\, \lambda> \lambda_0\,, \end{equation*}

and thus conclude that $A_{0,m}\in\mathcal{H}(X_{1,m})$ with $e^{tA_{0,m}} = e^{tA_{0,1}}|_{X_{1,m}}$ for $t\ge 0$ . The fact that $A_{0,m} \in \mathcal{G}(X_{1,m},1,\omega_m)$ follows again by interpolation using Lemma 3.2.

4 The absorption semigroup

Let $a\in L_{\infty,loc}([0,\infty))$ be a nonnegative function and $m\ge 1$ . We recall the definition (1.4) of the Schrödinger operator $A_{a,m}$ on $X_{1,m}$ given by

\begin{align*} \mathrm{dom}(A_{a,m}) & = \{ f\in X_{1,m}\ :\ f\in \mathrm{dom}(A_{0,m})\,, \ af \in X_{1,m}\}\,, \\ A_{a,m}f & = f'' - a f\,, \qquad f\in \mathrm{dom}(A_{a,m})\,. \end{align*}

The main result of this section is that $A_{a,m}\in \mathcal{G}_+(X_{1,m},1,\omega_m)\cap \mathcal{H}(X_{1,m})$ , the parameter $\omega_m$ being defined in Proposition 3.1. The proof relies on [Reference Arendt and Batty3].

Proposition 4.1 Assume that a satisfies (1.2) and let $m\ge 1$ . Then

\begin{equation*} A_{a,m}\in \mathcal{G}_+(X_{1,m},1,\omega_m)\cap \mathcal{H}(X_{1,m})\,. \end{equation*}

Moreover, $e^{tA_{a,m}} = e^{tA_{a,1}}|_{X_{1,m}}$ for $t\ge 0$ .

Proof. Let us recall that $X_{1,m}$ is a Banach lattice with order-continuous norm, see [Reference Aliprantis and Burkinshaw1, Chapter 4] for a precise definition and a complete list of properties, and introduce

\begin{equation*} Y := L_1((0,\infty),(x+x^m)a(x)\mathrm{d}x)\,. \end{equation*}

We first observe that $\left( \mathrm{dom}(A_{0,m})\cap Y \right)^\perp = \{0\}$ , where the disjoint complement $F^\perp$ of a subset F of the vector lattice $X_{1,m}$ is given by

\begin{equation*} F^\perp := \{ g \in X_{1,m}\ :\ \min\{|f|,|g|\} = 0 \;\text{ for all }\; f\in F\}\,. \end{equation*}

Indeed, since $C_c^\infty((0,\infty))$ is a subset of $\mathrm{dom}(A_{0,m})\cap Y$ , we readily deduce that $g\equiv 0$ for $g\in \left( \mathrm{dom}(A_{0,m})\cap Y \right)^\perp$ . Consequently, we are in a position to apply [Reference Arendt and Batty3, Proposition 4.3] and conclude that there is an extension $\hat{A}_{a,m}\in \mathcal{G}_+(X_{1,m})$ of $A_{a,m}$ with domain $\mathrm{dom}(\hat{A}_{a,m})$ defined as follows: $f\in \mathrm{dom}(\hat{A}_{a,m})$ if and only if $f\in X_{1,m}$ and there exist $(f_n)_{n\ge 1}$ in $\mathrm{dom}(A_{0,m})$ and $g\in X_{1,m}$ such that

\begin{equation*} \lim_{n\to\infty} \left( \|f_n-f\|_{X_{1,m}} + \|A_{0,m}f_n - (a\wedge n) f_n + g \|_{X_{1,m}} \right) = 0\,. \end{equation*}

It first follows from Lemma 4.2 below that $\mathrm{dom}(\hat{A}_{a,m})=\mathrm{dom}(A_{a,m})$ and therefore $\hat{A}_{a,m}=A_{a,m}$ . Moreover, $0\le e^{t A_{a,m}}=e^{t \hat{A}_{a,m}}\le e^{t A_{0,m}}$ for $t\ge 0$ by [Reference Arendt and Batty3, p. 432]. Since $A_{0,m}\in\mathcal{G}_+(X_{1,m},1,\omega_m)$ due to Proposition 3.1, this ordering property, along with [Reference Banasiak and Arlotti7, Remark 2.68], implies

\begin{equation*} \| e^{t \hat{A}_{a,m}}\|_{\mathcal{L}(X_{1,m})}\le \| e^{t A_{0,m}}\|_{\mathcal{L}(X_{1,m})}\le e^{\omega_m t}\,,\qquad t\ge 0\,. \end{equation*}

Hence $\hat{A}_{a,m}\in \mathcal{G}_+(X_{1,m},1,\omega_m)$ . Finally, recalling that

\begin{equation*} (-\omega_m+A_{0,m})\in\mathcal{G}_+(X_{1,m},1,0)\cap \mathcal{H}(X_{1,m}) \end{equation*}

by Proposition 3.1, we infer from [Reference Arendt and Batty3, Theorem 6.1] that $A_{a,m}\in \mathcal{H}(X_{1,m})$ .

It remains to check that $\mathrm{dom}(\hat{A}_{a,m})=\mathrm{dom}(A_{a,m})$ . This property actually follows from the monotonicity of the Laplace operator $A_{0,m}$ and the multiplication $f\mapsto af$ .

Proof. Pick $f\in \mathrm{dom}(\hat{A}_{a,m})$ . Then there are a sequence $(f_n)_{n\ge 1}$ in $\mathrm{dom}(A_{0,m})$ and $g\in X_{1,m}$ such that

(4.1) \begin{equation} \lim_{n\to\infty} \left( \|f_n-f\|_{X_{1,m}} + \|g_n - g \|_{X_{1,m}} \right) = 0\,, \end{equation}

with $g_n := - A_{0,m}f_n + (a\wedge n) f_n$ for $n\ge 1$ . In particular,

\begin{equation*} \kappa := \sup_{n\ge 1}\left\{ \|f_n\|_{X_{1,m}} + \|g_n\|_{X_{1,m}} \right\} < \infty\,. \end{equation*}

Step 1. Let us first prove that $af\in X_{1,m}$ . We infer from Lemma 2.3 that, for $n\ge 1$ ,

\begin{align*} \|g_n\|_{X_m} & \ge \int_0^\infty x^m\, \mathrm{sign}(f_n(x))\, g_n(x)\ \mathrm{d}x \\[3pt] & \ge m \int_0^\infty x^{m-1}\ |f_n|'(x)\ \mathrm{d}x + \int_0^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x \\[3pt] & = - m(m-1) \|f_n\|_{X_{m-2}} + \int_0^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x\,. \end{align*}

In particular, we derive

(4.2) \begin{equation} \int_0^\infty x\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x\le \|g_n\|_{X_1}\le \kappa\,. \end{equation}

Next, if $m\ge 3$ , then it follows from Young’s inequality that

\begin{align*} \int_0^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x & \le m(m-1) \left[ \frac{m-3}{m-1} \|f_n\|_{X_m} + \frac{2}{m-1} \|f_n\|_{X_1} \right] + \|g_n\|_{X_m} \\[3pt] & \le [m(m-1)+1] \kappa\,. \end{align*}

Likewise, if $m\in (1,3)$ , then Lemma 2.1 implies that

\begin{align*} \int_0^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x & \le m(m-1) \left[ \frac{1}{m-1} \|f_n^{\prime\prime}\|_{X_1} + \|f_n\|_{X_1} \right] + \|g_n\|_{X_m} \\[3pt] & \le m \|(a\wedge n) f_n -g_n\|_{X_1} + [m(m-1)+1]\kappa \\[3pt] & \le m \sup_{l\ge 1} \{\|(a\wedge l) f_l\|_{X_1}\} + (m^2+1) \kappa\\[3pt] & \le(m^2+m+1) \kappa \,, \end{align*}

where the last inequality is due to (4.2). Thus, in all cases for $m\ge 1$ , we have shown that

\begin{equation*} \int_0^\infty (x+x^m)\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x \le(m^2+m+2) \kappa\,. \end{equation*}

Fixing $N\ge 1$ , we deduce from the previous estimate that, for all $n\ge N$ ,

\begin{equation*} \int_0^\infty (x+x^m)\ (a(x)\wedge N) |f_n(x)|\ \mathrm{d}x \le \int_0^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x \le (m^2+m+2) \kappa\,. \end{equation*}

We then let $n\to\infty$ and infer from (4.1) that

\begin{equation*} \int_0^\infty (x+x^m)\ (a(x)\wedge N) |f(x)|\ \mathrm{d}x \le (m^2+m+2) \kappa\,. \end{equation*}

Using Fatou’s lemma to let $N\to\infty$ , we conclude that $af\in X_{1,m}$ with

(4.3) \begin{equation} \|a f\|_{X_{1,m}} \le (m^2+m+2) \kappa\,. \end{equation}

Step 2. We next show that $((a\wedge n) f_n)_{n\ge 1}$ converges to af in $X_{1,m}$ . Let $\chi\in C^\infty((0,\infty))$ be such that $\chi(x)=1$ for $x> 2$ , $\chi(x)=0$ for $x\in (0,1)$ , and $\chi(x)\in [0,1]$ for $x\in [1,2]$ . Introducing $\chi_R(x) := \chi(x/R)$ for $x\in (0,\infty)$ and $R>1$ , we deduce from Lemma 2.3 (with $\ell(x) = x^m \chi_R(x)$ ) that

\begin{align*} \int_0^\infty x^m\ (a(x)\wedge n) \chi_R(x) |f_n(x)|\ \mathrm{d}x & + \int_0^\infty [x^m \chi_R^{\prime}(x) + m x^{m-1} \chi_R(x)] |f_n|'(x)\ \mathrm{d}x \\[3pt] & \le \int_0^\infty x\chi_R(x) \mathrm{sign}(f_n(x)) g_n(x)\ \mathrm{d}x \\[3pt] & \le \int_R^\infty x\ |g_n(x)|\ \mathrm{d}x\,. \end{align*}

Since

\begin{align*} & \int_0^\infty [x^m \chi_R^{\prime}(x) + m x^{m-1} \chi_R(x)] |f_n|'(x)\ \mathrm{d}x \\[3pt] & = - \int_0^\infty [x^m \chi_R^{\prime\prime}(x) + 2m x^{m-1} \chi_R^{\prime}(x) + m(m-1) x^{m-2} \chi_R(x)] |f_n(x)|\ \mathrm{d}x \\[3pt] & = - \int_0^\infty x^m\ |f_n(x)| \left[ \frac{1}{R^2} \chi''\left( \frac{x}{R} \right) + \frac{2m}{Rx} \chi'\left( \frac{x}{R} \right) + \frac{m(m-1)}{x^2} \chi\left( \frac{x}{R} \right) \right]\ \mathrm{d}x \end{align*}

and

\begin{equation*} \left| \chi''(y) + \frac{2m}{y}\chi'(y) + \frac{m(m-1)}{y^2} \chi(y)\right| \le 3m^2 \|\chi''\|_{L_\infty(0,\infty)}\,, \qquad y\in (0,\infty)\,, \end{equation*}

we further obtain

\begin{align*} \int_{2R}^\infty x^m\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x & \le \int_0^\infty x^m\ (a(x)\wedge n) \chi_R(x) |f_n(x)|\ \mathrm{d}x \\[3pt] & \le \sup_{l\ge 1} \left\{ \int_R^\infty x^m\ |g_l(x)|\ \mathrm{d}x \right\} + \frac{3m^2}{R^2} \|\chi''\|_{L_\infty(0,\infty)} \|f_n\|_{X_m} \\[3pt] & \le \sup_{l\ge 1} \left\{ \int_R^\infty x^m\ |g_l(x)|\ \mathrm{d}x \right\} + \frac{3m^2\kappa}{R^2} \|\chi''\|_{L_\infty(0,\infty)} \end{align*}

for $n\ge 1$ . Since

\begin{equation*} \lim_{R\to\infty} \sup_{l\ge 1}\left\{ \int_R^\infty (x+x^m)\ |g_l(x)|\ \mathrm{d}x \right\} = 0 \end{equation*}

by (4.1), we conclude

(4.4) \begin{equation} \lim_{R\to\infty} \sup_{n\ge 1} \left\{ \int_{2R}^\infty (x+x^m)\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x \right\}= 0\,. \end{equation}

Now, let $R>1$ . Since $a\in L_\infty(0,2R)$ by (1.2), there is $n_R\ge 1$ such that $a(x)\wedge n=a(x)$ for $x\in (0,2R)$ and $n\ge n_R$ . Consequently, for $n\ge n_R$ ,

\begin{align*} \|(a\wedge n) f_n - af\|_{X_{1,m}} & \le \int_0^{2R} (x+x^m)\ a(x) |(f_n-f)(x)|\ \mathrm{d}x \\[3pt] & + \int_{2R}^\infty (x+x^m)\ (a(x)\wedge n) |f_n(x)|\ \mathrm{d}x \\[3pt] & + \int_{2R}^\infty (x+x^m)\ a(x) |f(x)|\ \mathrm{d}x \\[3pt] & \le \|a\|_{L_\infty(0,2R)} \|f_n-f\|_{X_{1,m}} \\[3pt] & + \sup_{l\ge 1} \left\{ \int_{2R}^\infty (x+x^m)\ (a(x)\wedge l) |f_l(x)|\ \mathrm{d}x \right\} \\ & + \int_{2R}^\infty (x+x^m)\ a(x) |f(x)|\ \mathrm{d}x \,. \end{align*}

We then pass to the limit as $n\to\infty$ and infer from (4.1) that

\begin{align*} \limsup_{n\to\infty} \|(a\wedge n) f_n - af\|_{X_{1,m}} &\le \sup_{l\ge 1} \left\{ \int_{2R}^\infty (x+x^m)\ (a(x)\wedge l) |f_l(x)|\ \mathrm{d}x \right\}\\[3pt] &\qquad + \int_{2R}^\infty (x+x^m)\ a(x) |f(x)|\ \mathrm{d}x\,. \end{align*}

We finally let $R\to\infty$ with the help of (4.3) and (4.4) and end up with

(4.5) \begin{equation} \lim_{n\to\infty} \|(a\wedge n) f_n - af\|_{X_{1,m}} = 0\,. \end{equation}

Step 3. We finally show that $f\in \mathrm{dom}( A_{0,m})$ . Indeed, it readily follows from (4.1) that $(f_n^{\prime\prime})_{n\ge 1}$ converges to f $^{\prime\prime}$ in the sense of distributions, while (4.1), (4.3), and (4.5) guarantee that $(f_n^{\prime\prime})_{n\ge 1}=((a\wedge n) f_n - g_n)_{n\ge 1}$ converges to $af -g$ in $X_{1,m}$ . Therefore, f $^{\prime\prime}$ belongs to $X_{1,m}$ with $f''=af-g$ and $(f_n^{\prime\prime})_{n\ge 1}$ converges to f $^{\prime\prime}$ in $X_{1,m}$ . Since $f_n(0)=0$ for $n\ge 1$ , this convergence along with Lemma 2.1 ensures that $f(0)=0$ and we have proved that $f\in \mathrm{dom}(A_{0,m})$ .

For further use, we show that the graph norm of $A_{a,m}$ in $X_{1,m}$ controls independently the diffusive and absorption terms in $X_{1,m}$ .

Lemma 4.3 Assume (1.2). For $f\in \mathrm{dom}(A_{a,1})$ ,

(4.6) \begin{equation} \frac{1}{3} \left( \|A_{0,1}f\|_{X_1} + \|af\|_{X_1} \right) \le \|A_{a,1}f\|_{X_1}\,. \end{equation}

Let $m>1$ . For $f\in \mathrm{dom}(A_{a,m})$ ,

(4.7) \begin{equation} \frac{1}{4(m+1)} \left( \|A_{0,m}f\|_{X_{1,m}} + \|af\|_{X_{1,m}} \right) - m \|f\|_{X_{1,m}} \le \|A_{a,m}f\|_{X_{1,m}}\,. \end{equation}

Proof. Let $f\in \mathrm{dom}(A_{a,1})$ and set $g:= -A_{a,1}f = -f''+af$ . It follows from Lemma 2.3 (with $\ell(x)=x$ ) that

\begin{equation*} \|g\|_{X_1} \ge \int_0^\infty x\ \mathrm{sign}(f(x)) g(x)\ \mathrm{d}x \ge \int_0^\infty |f|'(x)\ \mathrm{d}x + \|a f\|_{X_1} = \|a f\|_{X_1}\,. \end{equation*}

Consequently,

(4.8) \begin{equation} \|a f\|_{X_1} \le \|A_{a,1}f\|_{X_1} \;\text{ and }\; \|A_{0,1}f\|_{X_1} = \|A_{a,1}f + af\|_{X_1}\le 2 \|A_{a,1}f\|_{X_1}\,, \end{equation}

from which we deduce (4.6).

Next, let $m>1$ and consider $f\in \mathrm{dom}(A_{a,m})$ . We set $g:= -A_{a,m}f = -f''+af$ and infer from Lemma 2.3 (with $\ell(x)=x^m$ ) that

\begin{align*} \|g\|_{X_m} & \ge \int_0^\infty x^m\ \mathrm{sign}(f(x)) g(x)\ \mathrm{d}x \ge m \int_0^\infty x^{m-1} |f|'(x)\ \mathrm{d}x + \|a f\|_{X_m} \\[3pt] & = \|a f\|_{X_m} - m(m-1) \|f\|_{X_{m-2}}\,. \end{align*}

Either $m\ge 3$ and it follows from Young’s inequality and the above inequality that

(4.9a) \begin{align} \|a f\|_{X_m} & \le \|g\|_{X_m} + m(m-1) \left( \frac{m-3}{m-1} \|f\|_{X_m} + \frac{2}{m-1} \|f\|_{X_1} \right) \nonumber \\[3pt] & \le \|g\|_{X_m} + m(m-3) \|f\|_{X_m} + 2m \|f\|_{X_1} \nonumber \\[3pt] & \le \|g\|_{X_m} + m^2 \|f\|_{X_{1,m}} \,. \end{align}

Or $m\in (1,3)$ and we infer from (2.2) and (4.8) that

(4.9b) \begin{align} \|a f\|_{X_m} & \le \|g\|_{X_m} + m(m-1) \left( \frac{1}{m-1} \|f''\|_{X_1} + \|f\|_{X_1} \right) \nonumber \\[3pt] & \le \|g\|_{X_m} + m \|A_{0,1}f\|_{X_1} + m(m-1) \|f\|_{X_1} \nonumber \\[3pt] & \le \|g\|_{X_m} + 2m \|A_{a,1}f\|_{X_1} + m^2 \|f\|_{X_1} \,. \end{align}

Collecting (4.8) and (4.9) leads us to

\begin{equation*} \|af\|_{X_{1,m}} \le (1+2m) \|A_{a,m}f\|_{X_{1,m}} + m^2 \|f\|_{X_{1,m}} \,, \end{equation*}

which in turn gives

\begin{equation*} \|A_{0,m}f\|_{X_{1,m}} \le \|A_{a,m}f\|_{X_{1,m}} + \|af\|_{X_{1,m}} \le 2(1+m) \|A_{a,m}f\|_{X_{1,m}} + m^2 \|f\|_{X_{1,m}}\,. \end{equation*}

Consequently,

\begin{equation*} \frac{1}{4(1+m)} \left( \|A_{0,m}f\|_{X_{1,m}} + \|af\|_{X_{1,m}} \right) \le \|A_{a,m}f\|_{X_{1,m}} + \frac{m^2}{2(m+1)} \|f\|_{X_{1,m}}\,, \end{equation*}

from which (4.7) follows.

5 The fragmentation-diffusion semigroup

We now consider the operator $\mathbb{A}_m=A_{a,m}+B_m$ , where we recall that the nonlocal operator $B_{m}$ on $X_{1,m}$ is defined by

\begin{align*} \mathrm{dom}(B_m) & = \{ f\in X_{1,m}\ : \ af \in X_{1,m}\}\,, \\ B_mf(x) & = \int_x^\infty a(y) b(x,y) f(y)\ \mathrm{d}y\,, \quad x\in (0,\infty)\,, \qquad f\in \mathrm{dom}(B_m)\,. \end{align*}

We first show that $B_m$ is $A_{a,m}$ -bounded in $X_{1,m}$ .

Lemma 5.1 Assume (1.2) and (1.3). Let $m\ge 1$ and consider a measurable function f on $(0,\infty)$ such that $af\in X_m$ . Then

(5.1) \begin{equation} \int_0^\infty x^m \left| \int_x^\infty a(y) b(x,y) f(y)\ \mathrm{d}y \right|\ \mathrm{d}x \le \|af\|_{X_m}\,. \end{equation}

In addition,

(5.2) \begin{equation} M_1(B_mf)=M_1(af)\,,\qquad f\in \mathrm{dom}(B_m)\,, \end{equation}

and $B_m$ is $A_{a,m}$ -bounded in $X_{1,m}$ .

Proof. We infer from (1.3) and Fubini–Tonelli’s theorem that

\begin{align*} \int_0^\infty x^m \int_x^\infty a(y) b(x,y) |f(y)|\ \mathrm{d}y\mathrm{d}x & = \int_0^\infty a(y) |f(y)| \int_0^y x^m b(x,y)\ \mathrm{d}x\mathrm{d}y \\[3pt] & \le \int_0^\infty y^{m-1} a(y) |f(y)| \int_0^y x b(x,y)\ \mathrm{d}x\mathrm{d}y = \|af\|_{X_m}\,, \end{align*}

from which (5.1) readily follows. Next, (5.2) is a straightforward consequence of (1.3) and Fubini’s theorem.

Finally, let $f\in \mathrm{dom}(A_{a,m}) \subset \mathrm{dom}(B_m)$ . By (4.7) and (5.1),

\begin{align*} \|B_mf \|_{X_{1,m}} & \le \int_0^\infty (x+x^m) \left| \int_x^\infty a(y) b(x,y) f(y)\ \mathrm{d}y \right|\ \mathrm{d}x \le \|a f\|_{X_{1,m}} \\[3pt] & \le 4 (m+1) \|A_{a,m}f\|_{X_{1,m}} + 4m(m+1) \|f\|_{X_{1,m}}\,, \end{align*}

so that $B_m$ is $A_{a,m}$ -bounded.

As already observed in the literature, see, e.g., [Reference Banasiak, Lamb and Laurençot8, Theorem 5.1.47 (c)], the inequality (1.8) implies that, for each $m>1$ , there is $\delta_m\in (0,1)$ such that

(5.3) \begin{equation} (1-\delta_m) y^m \ge \int_0^y x^m b(x,y)\ \mathrm{d}x\,, \qquad y\in (0,\infty)\,. \end{equation}

An immediate consequence of (5.3) is a strict domination of a f over $B_m f$ in $X_m$ .

Lemma 5.2 Assume (1.2), (1.3), and (1.8). Let $m>1$ and consider $f\in \mathrm{dom}(B_m)$ . Then

\begin{equation*} \|B_m f\|_{X_m} \le (1-\delta_m) \|af\|_{X_m}\,. \end{equation*}

Proof. It readily follows from (5.3) and Fubini’s theorem that

\begin{align*} \|B_m f\|_{X_m} & \le \int_0^\infty x^m \int_x^\infty a(y) b(x,y) |f(y)|\ \mathrm{d}y\mathrm{d}x = \int_0^\infty a(y) |f(y)| \int_0^y x^m b(x,y)\ \mathrm{d}x\mathrm{d}y \\[3pt] & \le (1-\delta_m) \int_0^\infty y^m a(y) |f(y)| \mathrm{d}y =(1-\delta_m) \|af\|_{X_m}\,. \end{align*}

We shall see next that the property (5.3) ensures that $B_m$ is a Miyadera perturbation of $A_{a,m}$ . Recall that a similar result is available for the fragmentation equation without diffusion [Reference Banasiak6].

Proposition 5.3 Let $m>1$ and assume (1.2), (1.3), and (1.8). Then there are $q_m\in (0,1)$ and $t_m>0$ such that

\begin{equation*} \int_0^{t_m} \| B_m e^{sA_{a,m}}f\|_{X_{1,m}}\ \mathrm{d}s \le q_m \|f\|_{X_{1,m}}\,, \qquad f\in \mathrm{dom}(A_{a,m}) \,. \end{equation*}

In particular, $B_m$ is a Miyadera perturbation of $A_{a,m}$ .

Proof. Consider $f\in \mathrm{dom}(A_{a,m})$ and set $F(t) := e^{tA_{a,m}}f$ for $t\ge 0$ . Owing to Proposition 4.1, we have

(5.4) \begin{equation} \|F(t)\|_{X_1} \le \|f\|_{X_1}\,, \quad \|F(t)\|_{X_{1,m}} \le e^{\omega_m t} \|f\|_{X_{1,m}}\,, \qquad t\ge 0\,. \end{equation}

In addition, F is a classical solution to

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} F - A_{a,m} F = 0\,, \quad t>0\,, \qquad F(0)=f\,, \end{equation*}

and we deduce from Lemma 2.3 (with $\ell(x)=x^m$ ) that

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} \|F\|_{X_m} + \|a F\|_{X_m} \le m(m-1) \|F\|_{X_{m-2}}\,, \qquad t\ge 0\,. \end{equation*}

Hence, after integration with respect to time,

(5.5) \begin{equation} \int_0^t \|a F(s)\|_{X_m}\ \mathrm{d}s \le \|f\|_{X_m} + m(m-1) \int_0^t \|F(s)\|_{X_{m-2}}\ \mathrm{d}s\,, \qquad t\ge 0\,. \end{equation}

Now, let $t>0$ . It follows from (1.3), (5.1) (with $m=1$ ), and Lemma 5.2 that

(5.6) \begin{equation} \int_0^t \| B_mF(s)\|_{X_{1,m}}\ \mathrm{d}s \le \int_0^t \|a F(s)\|_{X_1}\ \mathrm{d}s + (1-\delta_m) \int_0^t \|a F(s)\|_{X_m}\ \mathrm{d}s\,. \end{equation}

For $R>1$ , we infer from (1.2) and (5.4) that

\begin{align*} \int_0^t \|a F(s)\|_{X_1}\ \mathrm{d}s & \le \|a\|_{L_\infty(0,R)} \int_0^t \int_0^R x |F(s,x)|\ \mathrm{d}x\mathrm{d}s \\[3pt] & \qquad + R^{1-m} \int_0^t \int_R^\infty x^m a(x) |F(s,x)|\ \mathrm{d}x\mathrm{d}s \\[3pt] & \le \|a\|_{L_\infty(0,R)} \int_0^t \|F(s)\|_{X_1}\ \mathrm{d}s + R^{1-m} \int_0^t \int_0^\infty x^m a(x) |F(s,x)|\ \mathrm{d}x\mathrm{d}s \\[3pt] & \le \|a\|_{L_\infty(0,R)} \|f\|_{X_1} t + R^{1-m} \int_0^t \| a F(s)\|_{X_m}\ \mathrm{d}s\,. \end{align*}

Combining (5.5), (5.6), and the above estimate with $R=R_m:=(\delta_m/2)^{1/(1-m)}$ gives

(5.7) \begin{align} \int_0^t \| B_mF(s)\|_{X_{1,m}}\ \mathrm{d}s & \le \|a\|_{L_\infty(0,R_m)} \|f\|_{X_1} t + \left( 1 - \frac{\delta_m}{2} \right) \int_0^t \| a F(s)\|_{X_m}\ \mathrm{d}s \nonumber \\[3pt] & \le \|a\|_{L_\infty(0,R_m)} \|f\|_{X_1} t + \left( 1 - \frac{\delta_m}{2} \right) \|f\|_{X_m} \nonumber \\[3pt] & \qquad + m(m-1)\left( 1 - \frac{\delta_m}{2} \right) \int_0^t \|F(s)\|_{X_{m-2}}\ \mathrm{d}s \nonumber \\[3pt] & \le \|a\|_{L_\infty(0,R_m)} \|f\|_{X_1} t + \left( 1 - \frac{\delta_m}{2} \right) \|f\|_{X_m} \nonumber \\[3pt] & \qquad + m(m-1) \int_0^t \|F(s)\|_{X_{m-2}}\ \mathrm{d}s\,. \end{align}

At this point, we handle the cases $m\ge 3$ and $m\in (1,3)$ in a different way. We first consider $m\ge 3$ . We use Young’s inequality, along with (5.4), to obtain

(5.8) \begin{align} m(m-1) \int_0^t \|F(s)\|_{X_{m-2}}\ \mathrm{d}s & \le m(m-1) \int_0^t \left[ \frac{m-3}{m-1} \|F(s)\|_{X_m} + \frac{2}{m-1} \|F(s)\|_{X_1} \right]\ \mathrm{d}s \nonumber \\[3pt] & \le \frac{m(m-3)}{1+\omega_m} \|f\|_{X_{1,m}} \left( e^{(1+\omega_m)t} - 1 \right) + 2m \|f\|_{X_1} t\,. \end{align}

Collecting (5.7) and (5.8) leads us to

\begin{align*} \int_0^t \| B_mF(s)\|_{X_{1,m}}\ \mathrm{d}s & \le \left[ \|a\|_{L_\infty(0,R_m)} t + \frac{m(m-3)}{1+\omega_m} \left( e^{(1+\omega_m)t} - 1 \right) + 2m t \right] \|f\|_{X_1} \\[3pt] & \qquad + \left[ 1 - \frac{\delta_m}{2} + \frac{m(m-3)}{1+\omega_m} \left( e^{(1+\omega_m)t} - 1 \right) \right] \|f\|_{X_m}\,. \end{align*}

We now pick $t_m>0$ such that

\begin{equation*} \left( \|a\|_{L_\infty(0,R_m)} + 2m \right) t_m \le 1 - \frac{\delta_m}{2} \quad \text{ and } \quad \frac{m(m-3)}{1+\omega_m} \left( e^{(1+\omega_m)t_m} - 1 \right)\le \frac{\delta_m}{4} \end{equation*}

and infer from the previous estimate (with $t=t_m$ ) that

\begin{equation*} \int_0^{t_m} \| B_mF(s)\|_{X_{1,m}}\ \mathrm{d}s \le \left( 1 - \frac{\delta_m}{4} \right) \|f\|_{X_{1,m}}\,. \end{equation*}

Recalling that $B_m$ is $A_{a,m}$ -bounded by Lemma 5.1, we have thus established that $B_m$ is a Miyadera perturbation of $A_{a,m}$ for $m\ge 3$ .

Let us now consider $m\in (1,3)$ . In that case, $m-2\in (-1,1)$ and it follows from Lemmas 2.1, 4.3, and (5.4) that, for $s\in (0,t)$ ,

\begin{align*} \|F(s)\|_{X_{m-2}} & \le \frac{2(3-m)^{(m-3)/2}}{m-1} \|F''(s)\|_{X_1}^{(3-m)/2} \|F(s)\|_{X_1}^{(m-1)/2} \\[3pt] & \le \frac{6(3-m)^{(m-3)/2}}{m-1} \|A_{a,1}F(s)\|_{X_1}^{(3-m)/2} \|f\|_{X_1}^{(m-1)/2}\,. \end{align*}

Owing to the analyticity of $\left( e^{tA_{a,m}} \right)_{t\ge 0}$ , see Proposition 4.1, we further infer from [Reference Pazy30, Theorem 2.5.2] that there is $C>0$ such that

\begin{equation*} \|A_{a,1} e^{s A_{a,1}} \|_{\mathcal{L}(X_1)} \le C \frac{e^s}{s} \le C \frac{e^t}{s}\,, \qquad s\in (0,t)\,. \end{equation*}

Combining the above two estimates gives

\begin{equation*} \|F(s)\|_{X_{m-2}} \le C(m) \|f\|_{X_1} e^{(3-m)t/2} s^{(m-3)/2}\,, \qquad s\in (0,t)\,. \end{equation*}

Hence, recalling (5.7),

\begin{align*} \int_0^t \| B_m F(s)\|_{X_{1,m}}\ \mathrm{d}s & \le \left[ \|a\|_{L_\infty(0,R_m)} t + C(m) e^{(3-m)t/2} t^{(m-1)/2} \right] \|f\|_{X_1} \\[3pt] & \qquad + \left( 1 - \frac{\delta_m}{2} \right) \|f\|_{X_m} \,. \end{align*}

We now choose $t_m>0$ such that

\begin{equation*} \|a\|_{L_\infty(0,R_m)} t_m + C(m) e^{(3-m)t_m/2} t_m^{(m-1)/2} \le 1 - \frac{\delta_m}{2} \end{equation*}

and deduce from the previous inequality (with $t=t_m$ ) that

\begin{equation*} \int_0^{t_m} \| B_m F(s)\|_{X_{1,m}}\ \mathrm{d}s \le \left( 1 - \frac{\delta_m}{2} \right) \|f\|_{X_{1,m}} \,. \end{equation*}

Consequently, using again Lemma 5.1, $B_m$ is also a Miyadera perturbation of $A_{a,m}$ when $m\in (1,3)$ .

We are now in a position to prove the first two statements in Theorem 1.1 for the operator $\mathbb{A}_m=A_{a,m}+B_m$ :

Proof of Theorem 1.1(a)-(b). We handle the cases $m=1$ and $m>1$ separately.

1. (a) If $m=1$ , then $A_{a,1}\in \mathcal{G}_+(X_1,1,0)$ by Proposition 4.1, so that it generates a substochastic semigroup in $X_1$ . Moreover, $\mathrm{dom}(A_{a,1})\subset \mathrm{dom}(B_1)$ and $B_1$ is obviously positive due to the nonnegativity of a and b. Also, for $f\in \mathrm{dom}(A_{a,1})$ ,

   \begin{equation*} M_1(A_{a,1}f+B_1f) = - \int_0^\infty f'(x)\ \mathrm{d}x - M_1(af) + M_1(B_1f) = 0 \end{equation*}
   by Lemma 2.1, (5.2), and the Dirichlet boundary condition. Consequently, we infer from [Reference Voigt31] and [Reference Banasiak, Lamb and Laurençot8, Theorem 4.9.16] that there is an extension $\tilde{\mathbb{A}}_1\in \mathcal{G}_+(X_1,1,0)$ of $\mathbb{A}_1$ .

2. (b) Let $m>1$ . Since $A_{a,m}\in\mathcal{H}(X_{1,m})$ by Proposition 4.1 and $B_m$ is a Miyadera perturbation of $A_{a,m}$ by Proposition 5.3, it follows from [Reference Engel and Nagel14, Corollary III.3.16 & Exercise III.3.17] that $\mathbb{A}_m=A_{a,m}+B_m\in \mathcal{H}(X_{1,m})$ with $\mathrm{dom}(\mathbb{A}_m) = \mathrm{dom}(A_{a,m})$ . Note that $D(A_{a,m})$ and $D(\mathbb{A}_m)$ are both Banach spaces and that $D(A_{a,m})$ is continuously embedded in $D(\mathbb{A}_m)$ , since $B_m$ is $A_{a,m}$ -bounded in $X_{1,m}$ according to Lemma 5.1. Consequently, $D(\mathbb{A}_m)\doteq D(A_{a,m})$ by the open mapping theorem.

We now check the positivity of $\left( e^{t\mathbb{A}_m} \right)_{t\ge 0}$ , bearing in mind that we already know from Proposition 4.1 that $A_{a,m}$ is resolvent positive. Pick $\lambda>0$ sufficiently large. Then $\lambda - \mathbb{A}_m$ is invertible with inverse given by

\begin{align*} (\lambda-\mathbb{A}_m)^{-1}&=(\lambda-A_{a,m}-B_m)^{-1} = (\lambda-A_{a,m})^{-1} \left( 1 - B_m (\lambda - A_{a,m})^{-1} \right)^{-1} \\[3pt] & = (\lambda-A_{a,m})^{-1} \sum_{j=0}^\infty \left[ B_m (\lambda - A_{a,m})^{-1} \right]^j\,, \end{align*}

where the Neumann series converges since $B_m$ is a Miyadera perturbation of $A_{a,m}$ , see the proof of [Reference Engel and Nagel14, Theorem III.3.14]. Now, $B_m$ is obviously a positive operator on $X_{1,m}$ due to the nonnegativity of a and b, and the positivity of $(\lambda-\mathbb{A}_m)^{-1}$ directly follows from the above identity.

Finally, as in the proof of (a), we have $M_1(\mathbb{A}_m f) = 0$ for any $f\in \mathrm{dom}(A_{a,m})$ by Lemma 2.1, (5.2), and the Dirichlet boundary condition, so that (1.9) immediately follows.

Proof of Proposition 1.2(a). Let $m\ge 1$ . The operator $A_{a,m}$ belongs to $\mathcal{G}_+(X_{1,m})\cap \mathcal{H}(X_{1,m})$ by Proposition 4.1. Since $a\in L_\infty(0,\infty)$ and b satisfies (1.3), the operator $B_m$ is a positive bounded operator on $X_{1,m}$ . On the one hand, it now follows from well-known perturbation results that $\mathbb{A}_m=A_{a,m}+B_m$ belongs to $\mathcal{H}(X_{1,m})$ , see [Reference Pazy30, Theorem 3.2.1]. On the other hand, the same argument as in the proof of Theorem 1.1(b) ensures the positivity of $\left( e^{t\mathbb{A}_m} \right)_{t\ge 0}$ . Finally, for $m=1$ , it readily follows from [Reference Banasiak, Lamb and Laurençot8, Proposition 4.9.16] that $\mathbb{A}_1=\tilde{\mathbb{A}}_1\in \mathcal{G}_+(X_1,1,0)$ , thereby completing the proof.

6 Immediate compactness of the semigroup

We now turn to compactness properties of the semigroup $(e^{t\mathbb{A}_m})_{t\ge 0}$ for $m>1$ as stated in Theorem 1.1 (c). To avoid loss of compactness for large sizes, we further require a to diverge to infinity for large sizes, thus excluding bounded overall fragmentation rates.

Proof. Recall that the relation $D(A_{a,m})\doteq D(\mathbb{A}_m)$ is established in the proof of Theorem 1.1 (b). Let $(f_n)_{n\ge 1}$ be a bounded sequence in $D (A_{a,m})$ . According to Lemma 2.1 and Lemma 4.3, there is $C>0$ such that

(6.1a) \begin{align} \sup_{x\ge 0} \left\{ |f_n(x)| + x|f_n^{\prime}(x)| \right\} & \le C\,, \qquad n\ge 1\,, \end{align}

(6.2b) \begin{align} \|f_n\|_{X_{1,m}} + \|f_n^{\prime\prime}\|_{X_1} + \|a f_n\|_{X_m} & \le C\,, \qquad n\ge 1\,. \end{align}

On the one hand, we infer from (6.1a) and Arzelà–Ascoli’s theorem that $(f_n)_{n\ge 1}$ is relatively compact in $C([1/R,R])$ for each $R>1$ . There are thus a subsequence $(f_{n_j})_{j\ge 1}$ and $f\in C((0,\infty))$ such that

(6.2) \begin{equation} \lim_{j\to\infty} f_{n_j}(x) = f(x)\,, \qquad x\in (0,\infty)\,. \end{equation}

On the other hand, it follows from (6.1) that, if $R>1$ and E is a measurable subset of $(0,\infty)$ , then, for $n\ge 1$ ,

(6.3) \begin{align} \int_E (x+x^m) |f_n(x)|\ \mathrm{d}x & \le \int_{E\cap (0,R)} (x+x^m) |f_n(x)|\ \mathrm{d}x + \int_R^\infty (x+x^m) |f_n(x)|\ \mathrm{d}x \nonumber \\[3pt] & \le (R+R^m) \|f_n\|_{L_\infty(0,\infty)} |E\cap (0,R)| \nonumber \\[3pt] & \qquad + \dfrac{2}{\inf_{x\ge R} \{a(x)\}} \int_R^\infty x^m\ a(x) |f_n(x)|\ \mathrm{d}x \nonumber \\[3pt] & \le 2 C R^m |E\cap (0,R)| + \dfrac{2C}{\inf_{x\ge R} \{a(x)\}}\,. \end{align}

A first consequence of (6.3) with $E=(R,\infty)$ is that

\begin{equation*} \sup_{n\ge 1} \int_R^\infty (x+x^m) |f_n(x)|\ \mathrm{d}x \le \dfrac{2C}{\inf_{x\ge R} \{a(x)\}}\,, \end{equation*}

from which we deduce by (1.10) that

(6.4) \begin{equation} \lim_{R\to\infty} \sup_{n\ge 1} \int_R^\infty (x+x^m) |f_n(x)|\ \mathrm{d}x = 0\,. \end{equation}

We next infer from (6.3) that, for $\delta>0$ ,

\begin{equation*} \eta(\delta) := \sup\left\{ \int_E (x+x^m) |f_n(x)|\ \mathrm{d}x\ :\ n \ge 1\,, \ E\in \mathcal{B}((0,\infty))\,, \ |E|\le\delta \right\} \end{equation*}

satisfies

\begin{equation*} 0 \le \eta(\delta) \le 2 C R^m \delta + \dfrac{2C}{\inf_{x\ge R} \{a(x)\}} \end{equation*}

for all $R>1$ . Therefore,

\begin{equation*} \limsup_{\delta\to 0} \eta(\delta) \le \dfrac{2C}{\inf_{x\ge R} \{a(x)\}} \end{equation*}

for all $R>1$ and we use once more (1.10) to conclude that

(6.5) \begin{equation} \lim_{\delta\to 0} \eta(\delta) = 0\,. \end{equation}

Gathering (6.4) and (6.5) implies that the sequence $(f_n)_{n\ge 1}$ is uniformly integrable in $X_{1,m}$ and thus weakly compact in $X_{1,m}$ by Dunford–Pettis’ theorem. This just established weak compactness in $X_{1,m}$ , along with the pointwise convergence (6.2) and Vitali’s theorem, see [Reference Fonseca and Leoni19, Theorem 2.24] for instance, entails that $(f_{n_j})_{j\ge 1}$ converges to f in $X_{1,m}$ , thereby completing the proof.

We are now in a position to finish off the proof of Theorem 1.1.

Proof of Theorem 1.1(c). Let $m>1$ . By Lemma 6.1, $(\lambda - \mathbb{A}_m)^{-1}$ is compact for $\lambda>0$ large enough and, since $m>1$ , the analyticity of $\left( e^{t\mathbb{A}_m} \right)_{t\ge 0}$ implies that it is continuous with respect to the operator norm for positive times [Reference Pazy30, Lemma 2.4.2]. We now may apply [Reference Pazy30, Theorem 2.3.3] to conclude that $\left( e^{t\mathbb{A}_m} \right)_{t\ge 0}$ is immediately compact.

The compactness result of Lemma 6.1 is not valid under the sole assumption (1.2) on a. In particular, we show that it fails when a is bounded.

Proof. Let $\varphi\in C_c^\infty(\mathbb{R})$ be such that $0\le \varphi\le 1$ , $\mathrm{supp}\ \varphi\subset [-1,1]$ , and $\|\varphi\|_{L_1(\mathbb{R})}=1$ . We fix $m\ge 1$ and set

\begin{equation*} \varphi_n(x) := \frac{1}{n+n^m} \varphi(x-n)\,, \qquad x\in (0,\infty)\,, \quad n\ge 1\,. \end{equation*}

Straightforward computations show that

\begin{align*} \|\varphi_n\|_{X_{1,m}} + \|a \varphi_n\|_{X_{1,m}} & \le \left( 1 + \|a\|_{L_\infty(0,\infty)} \right) \|\varphi_n\|_{X_{1,m}} \\[3pt] & \le \left( 1 + \|a\|_{L_\infty(0,\infty)} \right) \left( 1 + \int_{-1}^1 \frac{1+m2^m n^{m-1}}{n+n^m} \varphi(y)\ \mathrm{d}y \right) \\ & \le (2 + m 2^m) \left( 1 + \|a\|_{L_\infty(0,\infty)} \right) \end{align*}

and

\begin{align*} \|\varphi_n^{\prime\prime}\|_{X_{1,m}} \le \frac{n+1 + (n+1)^m}{n+n^m} \int_{-1}^1 |\varphi''(y)|\ \mathrm{d}y \le 2^m \|\varphi''\|_{L_1(\mathbb{R})}\,, \end{align*}

as well as

(6.6) \begin{equation} \lim_{n\to\infty} \|\varphi_n\|_{X_{1,m}} = 1\,, \qquad \lim_{n\to\infty} \|\varphi_n\|_{L_\infty(0,\infty)} = 0\,. \end{equation}

Therefore, the sequence $(\varphi_n)_{n\ge 1}$ is bounded in $D(A_{a,m})$ but cannot converge in $X_{1,m}$ due to (6.6).

Proof of Proposition 1.2(b). Let $m\ge 1$ and $a\in L_\infty(0,\infty)$ . Since $D(\mathbb{A}_m)$ is not compactly embedded in $X_{1,m}$ by Lemma 6.2, the resolvent $(\lambda-\mathbb{A}_m)^{-1}$ is not compact. Hence, [Reference Pazy30, Theorem 2.3.3] implies that the semigroup $(e^{t\mathbb{A}_m})_{t\ge 0}$ is not compact.

7 Steady states and convergence

Throughout this section, we assume that a and b satisfy (1.2), (1.3), (1.8), and (1.10) and that $a>0$ and $b>0$ .

We begin with the construction of stationary solutions with the help of Schauder’s fixed point theorem.

Lemma 7.1 There is a unique nonnegative

\begin{equation*} \psi_1\in \bigcap_{r\ge 1} \mathrm{dom}(\mathbb{A}_r) \end{equation*}

such that $M_1(\psi_1)=1$ and $\mathrm{ker}(\mathbb{A}_m) = \mathrm{ker}(\mathbb{A}_m^2) = \mathbb{R}\psi_1 = \{ r\psi_1\ :\ r\in\mathbb{R}\}$ for all $m\ge 1$ .

Proof. We split the proof into three steps.

Step 1. The uniqueness of a solution $\psi\in \mathrm{dom}(\mathbb{A}_1)$ to $\mathbb{A}_1\psi=0$ satisfying $M_1(\psi)=1$ relies on the dissipativity properties of $\mathbb{A}_1$ in $X_1$ and can be shown exactly as in the proofs of [Reference Escobedo, Mischler and Rodriguez Ricard15, Lemma 3.5] and [Reference Laurençot25, Proposition 3], to which we refer.

Step 2. We now turn to the existence part. Let $m\ge 3$ and consider $f\in X_{1,m}^+$ satisfying $M_1(f)=1$ . Setting $F(t) := e^{t\mathbb{A}_m}f$ for $t\ge 0$ , it readily follows from Theorem 1.1 (b) that

(7.1) \begin{equation} F(t)\ge 0 \;\text{ and }\; M_1(F(t))=1\,, \qquad t\ge 0\,. \end{equation}

Next, by (5.3) and Fubini’s theorem,

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) & = - m \int_0^\infty x^{m-1} \partial_x F(t,x)\ \mathrm{d}x - \int_0^\infty x^m a(x) F(t,x)\ \mathrm{d}x \\[3pt] & \qquad + \int_0^\infty a(y) F(t,y) \int_0^y x^m b(x,y)\ \mathrm{d}x\mathrm{d}y \\[3pt] & \le m(m-1) M_{m-2}(F(t)) - \delta_m M_m(aF(t))\,. \end{align*}

Owing to (1.10), there is $x_*>0$ such that $a(x)\ge 1$ for $x\ge x_*$ . Consequently, using (7.1),

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) & + \delta_m M_m(F(t)) \\[3pt] & \le \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) + \delta_m \int_{x_*}^\infty x^m F(t,x)\ \mathrm{d}x + \delta_m \int_0^{x_*} x^m F(t,x)\ \mathrm{d}x \\[3pt] & \le \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) + \delta_m M_m(aF(t)) + \delta_m x_*^{m-1} \int_0^{x_*} x F(t,x)\ \mathrm{d}x \\[3pt] & \le m(m-1) M_{m-2}(F(t)) + \delta_m x_*^{m-1} \,. \end{align*}

Since $m\ge 3$ , we now deduce from Young’s inequality that

\begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) + \delta_m M_m(F(t)) & \le \frac{\delta_m}{2} M_m(F(t)) + 2m \left( \frac{2m(m-3)}{\delta_m} \right)^{(m-3)/2} M_1(F(t)) \\[3pt] & \qquad + \delta_m x_*^{m-1} \,. \end{align*}

Hence, by (7.1),

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}t} M_m(F(t)) + \frac{\delta_m}{2} M_m(F(t)) \le 2m \left( \frac{2m(m-3)}{\delta_m} \right)^{(m-3)/2} + \delta_m x_*^{m-1}=: \frac{\delta_m \mu_m}{2} \end{equation*}

for $t\ge 0$ . After integration with respect to time, we conclude that

(7.2) \begin{equation} M_m(F(t)) \le \max\left\{ M_m(f) , \mu_m \right\}\,, \qquad t\ge 0\,. \end{equation}

Now, introducing

\begin{equation*} \mathcal{C}_m := \{ f\in X_{1,m}^+\ :\ M_1(f)=1\,, \ M_m(f) \le \mu_m\}\,, \end{equation*}

which is a closed convex subset of $X_{1,m}$ , an immediate consequence of (7.1) and (7.2) is that

\begin{equation*} e^{t\mathbb{A}_m}\mathcal{C}_m \subset \mathcal{C}_m\,, \qquad t\ge 0\,. \end{equation*}

Owing to the compactness of $e^{t\mathbb{A}_m}$ in $X_{1,m}$ for all $t>0$ , see Theorem 1.1(c), we argue as in the proofs of [Reference Amann2, Theorem 22.13] and [Reference Gamba, Panferov and Villani20, Theorem 5.2] to deduce from Schauder’s fixed point theorem that there is $\psi_m\in \mathcal{C}_m$ such that

\begin{equation*} e^{t\mathbb{A}_m}\psi_m = \psi_m\,, \qquad t\ge 0\,. \end{equation*}

Equivalently, $\mathbb{A}_m\psi_m=0$ and we have thus shown the existence of a stationary solution to (1.1a) for $m\ge 3$ . Obviously, $\psi_3$ also belongs to $\mathrm{dom}(\mathbb{A}_m)$ and satisfies $\mathbb{A}_m\psi_3=0$ for any $m\in [1,3)$ . Thus, there is at least one stationary solution $\psi_m$ to (1.1) for any $m\ge 1$ . Obviously, $\psi_m\in \mathrm{dom}(\mathbb{A}_1)$ solves $\mathbb{A}_1\psi_m=0$ for every $m\ge 1$ , and we infer from Step 1 that $\psi_m=\psi_1$ for every $m\ge 1$ .

Step 3. We finally identify $\mathrm{ker}(\mathbb{A}_m^2)$ . To this end, let $f\in \mathrm{ker}(\mathbb{A}_m^2)$ . Then $\mathbb{A}_m f$ belongs to $\mathrm{ker}(\mathbb{A}_m)$ , so that Step 2 implies that there is $\mu\in\mathbb{C}$ such that $\mathbb{A}_m f = \mu \psi_1$ . Therefore,

\begin{equation*} \mu = \mu M_1(\psi_1) = M_1(\mathbb{A}_m f)=0\,. \end{equation*}

Hence, $f\in \mathrm{ker}(\mathbb{A}_m)$ .

We now supply refined information on the spectrum of $\mathbb{A}_m$ for $m>1$ .

Lemma 7.2 Let $m>1$ . The spectrum $\sigma(\mathbb{A}_m)$ of $\mathbb{A}_m$ only consists of isolated eigenvalues and satisfies

(7.3) \begin{equation} \sigma(\mathbb{A}_m) \subset \{0\} \cup \{ \lambda\in \mathbb{C}\ :\ \textsf{Re}\, \lambda < -\varepsilon_m \} \end{equation}

for some $\varepsilon_m>0$ . Moreover, the spectral bound $s(\mathbb{A}_m)=0$ is a simple eigenvalue of $\mathbb{A}_m$ .

Proof. Owing to the immediate compactness of $(e^{t\mathbb{A}_m})_{t\ge 0}$ , see Theorem 1.1 (c), and [Reference Engel and Nagel14, Corollary V.3.2], the spectrum $\sigma(\mathbb{A}_m)$ only consists of isolated eigenvalues which are poles of the resolvent with finite algebraic multiplicity. Moreover, for any $r\in\mathbb{R}$ ,

(7.4) \begin{equation} \#\{\lambda\in \sigma(\mathbb{A}_m)\ :\ \textsf{Re}\,\lambda \ge r\} < \infty\,. \end{equation}

We next claim that $s(\mathbb{A}_m)=0$ . Indeed, since $\mathbb{A}_m \subset \mathbb{A}_1 \subset \tilde{\mathbb{A}}_1$ and $\tilde{\mathbb{A}}_1\in \mathcal{G}(X_1,1,0)$ by Theorem 1.1, any eigenvalue of $\mathbb{A}_m$ is also an eigenvalue of $\tilde{\mathbb{A}}_1$ and it follows from [Reference Pazy30, Corollary 1.3.6] that

\begin{equation*} \{ \lambda\in\mathbb{C}\ :\ \textsf{Re}\, \lambda > 0 \} \subset \rho(\tilde{\mathbb{A}}_1) \,. \end{equation*}

Consequently, any eigenvalue of $\mathbb{A}_m$ has a non-positive real part. Thus,

(7.5) \begin{equation} \sigma(\mathbb{A}_m)\subset \{ \lambda\in\mathbb{C}\ :\ \textsf{Re}\, \lambda \le 0 \}\,. \end{equation}

Since zero belongs to the spectrum of $\mathbb{A}_m$ by Lemma 7.1, we deduce from (7.5) that $s(\mathbb{A}_m)=0$ is a pole of the resolvent of $\mathbb{A}_m$ . Recalling that $(e^{t\mathbb{A}_m})_{t\ge 0}$ is a positive semigroup on the Banach lattice $X_{1,m}$ , it follows from [Reference Clément, Heijmans, Angenent, van Duijn and de Pagter12, Theorem 8.14] that $\sigma(\mathbb{A}_m)\cap i\mathbb{R}$ is either reduced to $\{0\}$ or contains infinitely many elements. The latter being ruled out by (7.4), we conclude that $\sigma(\mathbb{A}_m)\cap i\mathbb{R} = \{0\}$ . Since all eigenvalues are isolated, this last property ensures that there is $\varepsilon_m>0$ such that (7.3) holds true.

Finally, since $\mathrm{ker}(\mathbb{A}_m) = \mathrm{ker}(\mathbb{A}_m^2) = \mathbb{R}\psi_1$ by Lemma 7.1, zero is a simple eigenvalue of $\mathbb{A}_m$ according to [Reference Engel and Nagel14, Section IV.1.17].

Proof of Theorem 1.5. Let $m>1$ . From Lemma 7.1 we obtain the existence of a unique nonnegative

\begin{equation*} \psi_1\in \bigcap_{r\ge 1} \mathrm{dom}(\mathbb{A}_r) \end{equation*}

such that $M_1(\psi_1)=1$ and $\mathrm{ker}(\mathbb{A}_m) = \mathbb{R}\psi_1$ . We next infer from Lemma 7.2 that zero is a dominant eigenvalue of $\mathbb{A}_m$ and a first-order pole of its resolvent with residue P, where $P\in\mathcal{L}(X_{1,m})$ denotes the spectral projection onto $\mathrm{ker}(\mathbb{A}_m)$ and is given by

(7.6) \begin{equation} Pf = \lim_{\lambda\to 0} \lambda (\lambda - \mathbb{A}_m)^{-1}f\,, \qquad f\in X_{1,m}\,, \end{equation}

see, e.g., [Reference Engel and Nagel14, Section IV.1.17]. It then follows from [Reference Engel and Nagel14, Corollary V.3.3] that there are $N_m\ge 1$ and $\nu_m>0$ such that

(7.7) \begin{equation} \| e^{t\mathbb{A}_m} - P \|_{\mathcal{L}(X_{1,m})} \le N_m e^{-\nu_m t}\,, \qquad t\ge 0\,. \end{equation}

It only remains to identify the spectral projection P. Introducing $g_\lambda := \lambda (\lambda - \mathbb{A}_m)^{-1}f$ for $f\in X_{1,m}$ , we have

\begin{equation*} \lambda f = \lambda g_\lambda - \mathbb{A}_m g_\lambda\,, \end{equation*}

from which we readily deduce that $M_1(f) = M_1(g_\lambda)$ . Therefore, (7.6) implies $M_1(Pf) = M_1(f)$ . Since $Pf\in \mathbb{R}\psi_1$ and $M_1(\psi_1)=1$ , we conclude that

\begin{equation*} P f=M_1(f)\psi_1\,,\quad f\in X_{1,m}\,. \end{equation*}

Recalling (7.7), the above identity completes the proof of Theorem 1.5.

8 Stationary solutions revisited

We now prove the existence of a stationary solution to (1.1) when the overall fragmentation rate a may be bounded for large sizes but does not decay to zero. Specifically, we assume that a satisfies (1.12); that is,

\begin{equation*} \alpha := \frac{1}{2}\liminf_{x\to\infty} a(x) \in (0,\infty)\,.\end{equation*}

Proof of Proposition 1.6. As in Lemma 7.1, the proof of the uniqueness assertion in Proposition 1.6 relies on the dissipativity properties of $\mathbb{A}_1$ in $X_1$ and can be shown exactly as in the proofs of [Reference Escobedo, Mischler and Rodriguez Ricard15, Lemma 3.5] and [Reference Laurençot25, Proposition 3], to which we refer.

As for the existence assertion, we employ a compactness method. Let $n\ge 1$ . We set $a_n(x) := a(x) +x/n$ for $x>0$ and, for $m\ge 1$ , we denote the operators $B_m$ and $\mathbb{A}_m$ with $a_n$ instead of a by $B_{m,n}$ and $\mathbb{A}_{m,n}$ , respectively. Since $a_n(x)\to \infty$ as $x\to\infty$ , we infer from Lemma 7.1 that there is a unique nonnegative

\begin{equation*} \psi_{1,n}\in \bigcap_{r\ge 1} \mathrm{dom}(\mathbb{A}_{r,n}) \end{equation*}

such that $M_1(\psi_{1,n})=1$ and $\mathrm{ker}(\mathbb{A}_{m,n}) = \mathbb{R}\psi_{1,n}$ for all $m\ge 1$ . In particular, given $m>3$ , the function $\psi_{1,n}$ belongs to $\mathrm{dom}(A_{0,m})$ with $a_n \psi_{1,n}\in X_{1,m}$ and solves

(8.1) \begin{equation} - \psi_{1,n}^{\prime\prime} + a_n \psi_{1,n} = B_{m,n}\psi_{1,n} \;\text{ in }\; (0,\infty)\,, \qquad \psi_{1,n}(0)=0\,. \end{equation}

It follows from (8.1), Lemma 5.2, and Young’s inequality that, for $\varepsilon>0$ ,

\begin{align*} M_m(a_n \psi_{1,n}) & = M_m(B_{m,n}\psi_{1,n}) - m \int_0^\infty x^{m-1} \psi_{1,n}^{\prime}(x)\ \mathrm{d}x \\[3pt] & \le (1-\delta_m) M_m(a_n \psi_{1,n}) + m(m-1) M_{m-2}(\psi_{1,n}) \\[3pt] & \le (1-\delta_m) M_m(a_n \psi_{1,n}) + m(m-3) \varepsilon M_{m}(\psi_{1,n}) + 2m \varepsilon^{(3-m)/2} M_1(\psi_{1,n})\,. \end{align*}

Hence,

(8.2) \begin{equation} \delta_m M_m(a_n \psi_{1,n}) \le m(m-3) \varepsilon M_{m}(\psi_{1,n}) + 2m \varepsilon^{(3-m)/2}\,. \end{equation}

Owing to (1.12), there is $x_*>0$ such that

(8.3) \begin{equation} a(x) \ge \alpha\,, \qquad x\ge x_*\,. \end{equation}

In view of (8.2) and (8.3), we obtain

\begin{align*} \alpha \delta_m M_m(\psi_{1,n}) & \le \alpha\delta_m x_*^{m-1} \int_0^{x_*} x \psi_{1,n}(x)\ \mathrm{d}x + \delta_m \int_{x_*}^\infty x^m a_n(x) \psi_{1,n}(x)\ \mathrm{d}x \\[3pt] & \le \alpha\delta_m x_*^{m-1} M_1(\psi_{1,n}) + m(m-3) \varepsilon M_{m}(\psi_{1,n}) + 2m \varepsilon^{(3-m)/2}\,. \end{align*}

Choosing $\varepsilon=\alpha\delta_m/2m(m-3)$ in the above inequality gives

\begin{equation*} \frac{\alpha \delta_m}{2} M_m(\psi_{1,n}) \le \alpha\delta_m x_*^{m-1} + 2m \left( \frac{\alpha\delta_m}{2m(m-3)} \right)^{(3-m)/2}\,. \end{equation*}

Therefore, there is a positive constant $c_1(m)$ depending only on a and m such that

(8.4) \begin{equation} M_m(\psi_{1,n}) \le c_1(m)\,, \qquad n\ge 1\,. \end{equation}

Several additional estimates can now be derived from (8.4). Indeed, it readily follows from (8.1), (8.2) (with $\varepsilon=1$ ), and Lemma 5.2 that, for $n\ge 1$ ,

(8.5) \begin{align} \|\psi_{1,n}^{\prime\prime}\|_{X_m} + \|a_n \psi_{1,n}\|_{X_m} + \| B_{m,n}\psi_{1,n}\|_{X_m} & \le 2 \left( \|a_n \psi_{1,n}\|_{X_m} + \| B_{m,n}\psi_{1,n}\|_{X_m} \right) \nonumber \\[3pt] & \le 2(2-\delta_m) \|a_n \psi_{1,n}\|_{X_m} \le 4 M_m(a_n \psi_{1,n}) \nonumber \\[3pt] & \le \frac{4m(m-3)c_1(m)+8m}{\delta_m} =:c_2(m)\,. \end{align}

Similarly, by (1.2), (8.1), (8.4) (with $m=4$ ), and Lemma 5.1,

(8.6) \begin{align} & \|\psi_{1,n}^{\prime\prime}\|_{X_1} + \|a_n \psi_{1,n}\|_{X_1} + \| B_{m,n}\psi_{1,n}\|_{X_1} \nonumber\\[3pt] & \qquad\qquad \le 2 \left( \|a_n \psi_{1,n}\|_{X_1} + \| B_{m,n}\psi_{1,n}\|_{X_1} \right) \le 4 \|a_n \psi_{1,n}\|_{X_m} \nonumber \\[3pt] & \qquad\qquad \le 4 \left( 1 + \|a\|_{L_\infty(0,1)} \right) \int_0^1 x \psi_{1,n}(x)\ \mathrm{d}x + 4 \int_1^\infty x^4 a_n(x) \psi_{1,n}(x)\ \mathrm{d}x \nonumber \\[3pt] & \qquad\qquad \le 4 \left( 1 + \|a\|_{L_\infty(0,1)} \right) + 4 c_2(4)=:c_3 \end{align}

for $n\ge 1$ . Moreover, a straightforward consequence of (8.5), (8.6), and Hölder’s inequality is that (8.5) is also true for $m\in (1,3]$ with a suitable constant $c_2(m)$ .

We next claim that $(\psi_{1,n})_{n\ge 1}$ is relatively compact in $X_{1,m}$ for any $m\ge 1$ . To this end, we note that, thanks to (8.5), (8.6), Lemma 2.1, and Fubini’s theorem,

\begin{equation*} \|\psi_{1,n}\|_{X_0} \le \frac{1}{\sqrt{2}} \|\psi_{1,n}^{\prime\prime}\|_{X_1}^{1/2} \, \|\psi_{1,n}\|_{X_1}^{1/2} \le \sqrt{\frac{c_3}{2}} \end{equation*}

and

\begin{equation*} \|\psi_{1,n}\|_{X_1} + \|\psi_{1,n}^{\prime}\|_{X_1} \le 1 + \int_0^\infty x \int_x^\infty |\psi_{1,n}^{\prime\prime}(y)|\ \mathrm{d}y\mathrm{d}x \le 1 + \frac{\|\psi_{1,n}^{\prime\prime}\|_{X_2}}{2} \le 1+c_2(2) \end{equation*}

for $n\ge 1$ . Now, let $m\ge 2$ . In view of (8.5) and the above estimates, $(\psi_{1,n})_{n\ge 1}$ is a bounded sequence in $X_0\cap X_m\cap W_1^1((0,\infty),x\mathrm{d}x)$ and it follows from [Reference Banasiak, Lamb and Laurençot9, Proposition 7.2.2] that $(\psi_{1,n})_{n\ge 1}$ is relatively compact in $X_r$ for any $r\in (0,m)$ . As $m\ge 2$ is arbitrary, we conclude that there are $\psi_1\in \bigcap_{r>0} X_r$ and a subsequence $(\psi_{1,n_j})_{j\ge 1}$ of $(\psi_{1,n})_{n\ge 1}$ such that

(8.7) \begin{equation} \lim_{j\to\infty} \|\psi_{1,n_j} - \psi_1\|_{X_r} = 0 \quad\text{for all}\; r>0.\end{equation}

An immediate consequence of (8.7) and the properties of $(\psi_{1,n})_{n\ge 1}$ is that

\begin{equation*} \psi_1 \in X_1^+ \;\;\text{and}\;\; M_1(\psi_1)=1.\end{equation*}

Finally, let $m\ge 1$ . We observe that $\mathrm{dom}(\mathbb{A}_{m,n})\subset \mathrm{dom}(\mathbb{A}_m)$ for $n\ge 1$ (as $a_n\ge a$ ) and that (8.1) also reads

(8.8) \begin{equation} \mathbb{A}_m \psi_{1,n} = \mathcal{R}_n\,, \end{equation}

where

\begin{equation*} \mathcal{R}_n(x) := - \frac{x}{n} \psi_{1,n}(x) + \frac{1}{n} \int_x^\infty y b(x,y) \psi_{1,n}(y)\ \mathrm{d}y\,, \qquad x>0\,. \end{equation*}

It readily follows from (8.5) that

\begin{equation*} \|\mathcal{R}_n \|_{X_{1,m}} \le \frac{2}{n} \left( \|\psi_{1,n}\|_{X_2} + \|\psi_{1,n}\|_{X_{m+1}} \right) \le \frac{2}{n} \left( c_2(2) + c_2(m+1) \right) \,, \end{equation*}

so that

(8.9) \begin{equation} \lim_{n\to\infty} \|\mathcal{R}_n \|_{X_{1,m}} =0\,. \end{equation}

In view of (8.8) and (8.9), the sequence $(\mathbb{A}_m \psi_{1,n_j})_{j\ge 1}$ converges to zero in $X_{1,m}$ as $j\to\infty$ and, since $\mathbb{A}_m$ is closed on $X_{1,m}$ , we readily deduce from (8.7) that $\psi_1\in \mathrm{dom}(\mathbb{A}_m)$ and $\mathbb{A}_m\psi_1=0$ .