Proof Set $u = \log \varphi $ on $\mathbb {D}$ . Since $\varphi $ is radial and logarithmically subharmonic, u is radial and subharmonic, and u is not identically $-\infty $ .

We first show that $u(r)> -\infty $ for all $0<r<1$ . Suppose, for contradiction, that there exists $r_0\in (0,1)$ with $u(r_0)=-\infty $ . Because u is radial, $u(z)=u(|z|)=-\infty $ for every $|z|=r_0$ . Consequently,

$$\begin{align*}\frac{1}{2\pi}\int_{0}^{2\pi} u\bigl(r_0 e^{i\theta}\bigr)\,d\theta = -\infty. \end{align*}$$

On the other hand, u is subharmonic on $\mathbb {D}$ and $u\not \equiv -\infty $ . A standard property of subharmonic functions (see, e.g., [Reference RansfordRan95, Corollary 2.5.2]) states that if u is subharmonic on a domain $D\subset \mathbb {C}$ , $u\not \equiv -\infty $ , and $\{z:|z-w|\le \rho \}\subset D$ , then

$$\begin{align*}\frac{1}{2\pi}\int_{0}^{2\pi} u\bigl(w+\rho e^{i\theta}\bigr)\,d\theta> -\infty. \end{align*}$$

Applying this with $D=\mathbb {D}$ , $w=0$ , and $\rho =r_0$ yields

$$\begin{align*}\frac{1}{2\pi}\int_{0}^{2\pi} u\bigl(r_0 e^{i\theta}\bigr)\,d\theta> -\infty , \end{align*}$$

a contradiction. Hence, $u(r)>-\infty $ for every $r\in (0,1)$ .

We now prove the convexity of g. Fix $t_1,t_2>0$ and $\lambda \in (0,1)$ , and set

$$\begin{align*}t = (1-\lambda)t_1 + \lambda t_2. \end{align*}$$

Define

$$\begin{align*}r_1 = e^{-t_1/2},\qquad r_2 = e^{-t_2/2},\qquad r = e^{-t/2}. \end{align*}$$

Then

$$\begin{align*}r = r_1^{\,1-\lambda}\, r_2^{\,\lambda}. \end{align*}$$

Without loss of generality, assume $t_1<t_2$ , so that $r_1>r_2$ .

Consider the annulus $A = \{z\in \mathbb {C}: r_2<|z|<r_1\}$ . On A define

$$\begin{align*}f(z) = \frac{\log(r_1/|z|)}{\log(r_1/r_2)}\,u(r_2) + \frac{\log(|z|/r_2)}{\log(r_1/r_2)}\,u(r_1),\qquad z\in A. \end{align*}$$

Since $\log |z|$ is harmonic on A, f is harmonic on A. Moreover, f is radial and satisfies

$$\begin{align*}f(\zeta)=u(r_1)\;\;(|\zeta|=r_1),\qquad f(\zeta)=u(r_2)\;\;(|\zeta|=r_2). \end{align*}$$

The function $u-f$ is subharmonic on A because u is subharmonic and f is harmonic. For any boundary point $\zeta \in \partial A$ , we have

$$\begin{align*}\limsup_{A\ni z\to\zeta}\bigl(u(z)-f(z)\bigr)\le 0. \end{align*}$$

Indeed, if $|\zeta |=r_1$ , radiality gives $\limsup _{z\to \zeta }u(z)\le u(r_1)=f(\zeta )$ ; similarly, if $|\zeta |=r_2$ , then $\limsup _{z\to \zeta }u(z)\le u(r_2)=f(\zeta )$ . By the maximum principle for subharmonic functions,

$$\begin{align*}u(z)\le f(z)\qquad\text{for all }z\in A. \end{align*}$$

In particular, for any z with $|z|=r$ , we have $u(r)\le f(r)$ .

Using $r=r_1^{\,1-\lambda }r_2^{\,\lambda }$ , we compute

$$\begin{align*}\log\frac{r_1}{r} = \lambda\log\frac{r_1}{r_2},\qquad \log\frac{r}{r_2} = (1-\lambda)\log\frac{r_1}{r_2}. \end{align*}$$

Therefore,

$$\begin{align*}f(r) = \lambda\,u(r_2) + (1-\lambda)\,u(r_1). \end{align*}$$

Thus,

$$\begin{align*}u(r) \le (1-\lambda)u(r_1) + \lambda u(r_2). \end{align*}$$

Recalling that $r=e^{-t/2}$ , $r_1=e^{-t_1/2}$ , $r_2=e^{-t_2/2}$ , we obtain

$$\begin{align*}u\bigl(e^{-t/2}\bigr) \le (1-\lambda)\,u\bigl(e^{-t_1/2}\bigr) + \lambda\,u\bigl(e^{-t_2/2}\bigr), \end{align*}$$

that is,

$$\begin{align*}g\bigl((1-\lambda)t_1+\lambda t_2\bigr) \le (1-\lambda)g(t_1) + \lambda g(t_2). \end{align*}$$

Hence, g is convex on $(0,\infty )$ . This completes the proof.

Proof Since $f>0$ , the function $ k(t)=\log f(t) $ is well defined and is convex on $(0,\infty )$ . Define a convex function $\alpha $ on $(0,\infty )$ by

$$\begin{align*}\alpha(t)=k(t)+t=\log f(t)+t \end{align*}$$

and recall the right half-plane $ \mathbb H=\{w\in \mathbb C: \mathrm {Re}w>0\}, $ then the function $ U(w)=\alpha (\mathrm {Re} w) $ is subharmonic on $\mathbb H$ .

Next fix $z_0\in \mathbb D\setminus \{0\}$ . Choose a simply connected neighborhood $V\subset \mathbb D\setminus \{0\}$ of $z_0$ on which a holomorphic branch of the logarithm exists; denote it by $\mathrm {Log} z$ . Let

$$\begin{align*}F(z):=-2\mathrm{Log} z, \qquad z\in V. \end{align*}$$

Then F is holomorphic on V, and

$$\begin{align*}\mathrm{Re} F(z)=\mathrm{Re}(-2\mathrm{Log} z)=-2\log|z|>0, \qquad z\in V, \end{align*}$$

because $0<|z|<1$ . Hence, $F(V)\subset \mathbb H$ .

Since U is subharmonic on $\mathbb H$ and F is holomorphic, the composition $U\circ F$ is subharmonic on V. On the other hand, for $z\in V$ ,

$$\begin{align*}(U\circ F)(z) = \alpha(\mathrm{Re} F(z)) = \alpha(-2\log|z|). \end{align*}$$

By the definition of $\alpha $ ,

$$\begin{align*}\alpha(-2\log|z|) = \log f(-2\log|z|)-2\log|z|. \end{align*}$$

Since

$$\begin{align*}\varphi(z)=\frac{f(-2\log|z|)}{|z|^2}, \end{align*}$$

it follows that

$$\begin{align*}(U\circ F)(z)=\log\varphi(z). \end{align*}$$

Therefore, $\log \varphi $ is subharmonic on V. Because $z_0$ was arbitrary, we conclude that $\log \varphi $ is subharmonic on $\mathbb D\setminus \{0\}$ . Equivalently, $\varphi $ is logarithmically subharmonic on $\mathbb D\setminus \{0\}$ .

For the last assertion, assume that $\varphi $ is bounded above in a neighborhood of $0$ . Then $\log \varphi $ is bounded above in a punctured neighborhood of $0$ and is subharmonic on $\mathbb D\setminus \{0\}$ . By the removable singularity theorem for subharmonic functions, the function

$$\begin{align*}\widetilde u(z):= \begin{cases} \log\varphi(z), & z\neq 0,\\[4pt] \limsup\limits_{\zeta\to 0}\log\varphi(\zeta), & z=0, \end{cases} \end{align*}$$

is subharmonic on $\mathbb D$ . Setting

$$\begin{align*}\widetilde\varphi(z):=e^{\widetilde u(z)}, \end{align*}$$

we obtain a logarithmically subharmonic extension of $\varphi $ to the whole disk. This completes the proof.

Proof of Theorem 2.1

Necessity. Assume that there exists a radial logarithmically subharmonic weight $\omega $ on $\mathbb D$ such that

$$\begin{align*}K_\omega(z,\lambda)=S_\nu(z,\lambda), \qquad z,\lambda\in\mathbb D. \end{align*}$$

Then

$$\begin{align*}0<\int_0^1 r^{n+1}\omega(r) \mathrm{d} r<\infty \qquad \forall n\in\mathbb{N}. \end{align*}$$

By the coefficient formulas for the kernels $K_\omega $ and $S_\nu $ , it follows that for every $n\in \mathbb N$ ,

$$\begin{align*}\left(\int_0^1 \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r)\right) \left(2\int_0^1 r^{2n+1}\omega(r) \mathrm{d} r\right)=1. \end{align*}$$

Define the function $h:(0,\infty )\to (0,\infty )$ by

$$\begin{align*}h(t)=e^{-t}\omega(e^{-t/2}). \end{align*}$$

By Lemma 2.2, the function $t\mapsto \log \omega (e^{-t/2})$ is convex on $(0,\infty )$ . Hence,

$$\begin{align*}\log h(t)=\log\omega(e^{-t/2})-t \end{align*}$$

is also convex on $(0,\infty )$ , because subtracting an affine function preserves convexity. Thus h is logarithmically convex.

Next, by the change of variables $r=e^{-t/2}$ , we obtain, for each $n\in \mathbb N$ ,

$$\begin{align*}\int_0^\infty e^{-nt}h(t) \mathrm{d} t =\int_0^\infty e^{-nt}e^{-t}\omega(e^{-t/2}) \mathrm{d} t = 2\int_0^1 r^{2n+1}\omega(r) \mathrm{d} r. \end{align*}$$

Therefore,

$$\begin{align*}\left(\int_0^1 \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r)\right) \left(\int_0^\infty e^{-nt}h(t) \mathrm{d} t\right)=1, \qquad n\in\mathbb N. \end{align*}$$

It remains to verify the growth condition. Since $\log \omega $ is subharmonic on $\mathbb D$ , it is locally bounded above near $0$ . Hence, $\omega $ is bounded above in some neighborhood of $0$ . Therefore, for all sufficiently large t,

$$\begin{align*}e^t h(t)=\omega(e^{-t/2}) \end{align*}$$

is bounded, that is, there exists $T>0$ such that

$$\begin{align*}\sup_{t\ge T} e^t h(t)<\infty. \end{align*}$$

Sufficiency. Assume that there exists a logarithmically convex function $h:(0,\infty )\to (0,\infty )$ such that

$$\begin{align*}\left(\int_0^1 \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r)\right) \left(\int_0^\infty e^{-nt}h(t) \mathrm{d} t\right)=1, \qquad n\in\mathbb N, \end{align*}$$

and such that

$$\begin{align*}\sup_{t\ge T} e^t h(t)<\infty \end{align*}$$

for some $T>0$ .

For any point z with $0<|z|<1$ , define

$$\begin{align*}\omega(z)=\frac{h(-2\log|z|)}{|z|^2}. \end{align*}$$

By Lemma 2.3, the function $\omega $ is logarithmically subharmonic on $\mathbb D\setminus \{0\}$ , and the growth condition

$$\begin{align*}\sup_{t\ge T}e^t h(t)<\infty \end{align*}$$

implies that $\omega $ is bounded above near $0$ . Hence, $\omega $ extends to a logarithmically subharmonic function on $\mathbb D$ . Since the original function on $\mathbb D\setminus \{0\}$ is radial, the extension is radial as well. Since $h(t)>0$ for all $t>0$ , we have $\omega (r)>0$ for all $0<r<1$ . Hence

$$\begin{align*}\int_r^1 \omega(t) t \mathrm{d} t>0 \end{align*}$$

for every $0<r<1$ , so $L_a^2(\omega )$ is a reproducing kernel Hilbert space. We next show that

$$\begin{align*}\int_0^1 r\omega(r) \mathrm{d} r<\infty. \end{align*}$$

Taking $n=0$ , we get

$$\begin{align*}\nu([0,1])\int_0^\infty h(t) \mathrm{d} t=1. \end{align*}$$

Since $\nu ([0,1])>0$ , it follows that

$$\begin{align*}\int_0^\infty h(t) \mathrm{d} t=\frac{1}{\nu([0,1])}<\infty. \end{align*}$$

Using again the change of variables $t=-2\log r$ , we obtain

$$\begin{align*}\int_0^1 r\omega(r) \mathrm{d} r =\int_0^1 \frac{h(-2\log r)}{r} \mathrm{d} r =\frac{1}{2}\int_0^\infty h(t) \mathrm{d} t =\frac{1}{2 \nu([0,1])}<\infty. \end{align*}$$

Hence, for every $n\in \mathbb N$ ,

$$\begin{align*}\int_0^1 r^{2n+1}\omega(r) \mathrm{d} r\le\int_0^1 r \omega(r) \mathrm{d} r<\infty. \end{align*}$$

Moreover,

$$ \begin{align*} 2\int_0^1 r^{2n+1}\omega(r) \mathrm{d} r &=2\int_0^1 r^{2n-1} h(-2\log r) \mathrm{d} r\\ &=\int_0^\infty e^{-nt}h(t) \mathrm{d} t. \end{align*} $$

Hence, for each $n\in \mathbb N$ ,

$$\begin{align*}\left(\int_0^1 \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r)\right) \left(2\int_0^1 r^{2n+1}\omega(r) \mathrm{d} r\right)=1. \end{align*}$$

By the coefficient formulas for $K_\omega $ and $S_\nu $ , this implies

$$\begin{align*}K_\omega(z,\lambda)=S_\nu(z,\lambda), \qquad z,\lambda\in\mathbb D. \end{align*}$$

This completes the proof.

Proof of Proposition 3.1.

Let $\nu $ be a finite positive Borel measure on $[0,1]$ , and let $S_{\nu }$ be the corresponding Shimorin-type kernel. Suppose that

$$ \begin{align*}S_{\nu}(z,\lambda) = K_{\omega}(z,\lambda), \quad z, \lambda \in \mathbb{D},\end{align*} $$

for some measure $\omega $ on $[0,1)$ with $\omega ([r,1))> 0$ for all $0 < r < 1$ . Note that

$$\begin{align*}S_{\nu}(z,\lambda) = \sum_{n = 0}^{\infty} \left(\int_0^1 \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r)\right) (z\bar{\lambda})^n, \end{align*}$$

and

$$\begin{align*}K_{\omega}(z,\lambda) = \sum_{n = 0}^{\infty} \frac{1}{2\omega_{2n + 1}} (z\bar{\lambda})^n, \end{align*}$$

where $\omega _{2n + 1} = \int _0^1 r^{2n + 1} \mathrm {d} \omega (r)$ . Then we have

$$\begin{align*}\int_{0}^{1} \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) = \frac{1}{2 \omega_{2n + 1}}. \end{align*}$$

By the Lebesgue dominated convergence theorem,

$$\begin{align*}\lim_{n \to \infty} \omega_{2n + 1} = \lim_{n \to \infty} \int_0^1 r^{2n + 1} \mathrm{d} \omega(r) = 0. \end{align*}$$

Hence, by the monotone convergence theorem,

$$\begin{align*}\int_{0}^{1} \frac{1}{1 - r} \mathrm{d} \nu(r) = \lim_{n \to \infty} \int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) = \lim_{n \to \infty} \frac{1}{2 \omega_{2n + 1}} = \infty. \end{align*}$$

Conversely, consider the Hausdorff moment sequence of the measure $\nu $

$$\begin{align*}v_k = \int_0^1 r^k \mathrm{d} \nu(r), \quad k \in \mathbb{N}, \end{align*}$$

by Lemma 3.2, there exists a finite positive Borel measure $\mu $ on $[0,1]$ such that

$$\begin{align*}\int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) \cdot \int_0^1 r^n \mathrm{d} \mu(r) = 1 , \quad n \in \mathbb{N}. \end{align*}$$

Let $\psi (x) = \sqrt {x}$ . By a change of variables,

$$\begin{align*}\int_{0}^{1} r^n \mathrm{d} \mu(r) = \int_{0}^{1} t^{2n} \mathrm{d} (\psi_{*}\mu)(t) = \int_0^1 t^{2n + 1} \frac{1}{t} \mathrm{d} (\psi_{*} \mu)(t). \end{align*}$$

Then we can define a measure $\omega $ on $[0,1]$ by

(3.2) $$ \begin{align} \omega(E) = \frac{1}{2} \int_{E} \frac{1}{t} \mathrm{d} (\psi_{*}\mu)(t) = \frac{1}{2}\int_{\psi^{-1}(E)}r^{-1 / 2} \mathrm{d} \mu(r), \end{align} $$

where $E\subset [0,1]$ is a Borel set. Hence, we have

$$\begin{align*}2 \omega_{2n + 1} = \int_0^1 t^{2n + 1} \frac{1}{t} \mathrm{d} (\psi_{*}\mu)(t) = \int_0^1 r^n \mathrm{d} \mu(r) = \frac{1}{\int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r)}. \end{align*}$$

To complete the proof, it remains to show that

• • $\omega $ is a finite measure on $[0,1]$ , i.e., $\int _{0}^{1} r^{- 1/ 2} \mathrm {d} \mu (r) < \infty $ ;

• • $\omega (\{1\}) = 0$ , that is, $\omega $ is concentrated on $[0,1)$ ;

• • for all $0 < r < 1$ , $\omega ([r,1))> 0$ ;

• • $\omega $ is a $\widehat {\mathcal {D}}$ -weight.

Let f be the nonzero Bernstein function given by

$$\begin{align*}f(s) = \int_0^1 \frac{1 - t^{s}}{1 - t} \mathrm{d} \nu(t). \end{align*}$$

By Lemma 3.3, there is a $\sigma $ -finite positive Borel measure $\widetilde {\kappa }$ on $[0,1]$ such that

$$\begin{align*}\mathrm{d} \mu(t) = t\mathrm{d} \widetilde{\kappa}(t), \quad \text{and} \quad \int_{0}^{1} t^s \mathrm{d} \widetilde{\kappa}(t) = \frac{1}{f(s)}, \ s> 0. \end{align*}$$

Hence,

$$\begin{align*}\int_0^1 r^{-1 / 2}\mathrm{d} \mu(r) = \int_0^1 t^{1/2} \mathrm{d} \widetilde{\kappa}(t) = \frac{1}{f(1/2)} < \infty. \end{align*}$$

Next, to show $\omega (\{1\}) = 0$ , we only need to show $\mu (\{1\}) = 0$ (Indeed, $\omega (\{1\}) = \frac {1}{2} \mu (\{1\})$ ). If $\mu (\{1\}) \ne 0$ , then

$$\begin{align*}\int_0^1 r^{n} \mathrm{d} \mu(r) \geq \mu(\{1\})> 0. \end{align*}$$

Since $\int _{0}^{1} \frac {1}{1 - r}\mathrm {d} \nu (r) = \infty $ , we have

$$\begin{align*}\lim_{n\to\infty}\int_{0}^{1} \frac{1 - r^{n + 1}}{1 - r} \mathrm{d} \nu(r) = \infty.\end{align*}$$

This contradicts the fact that

$$\begin{align*}\int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) \int_{0}^{1} r^n \mathrm{d} \mu(r) = 1. \end{align*}$$

Finally, we show that for all $0 < r < 1$ ,

$$\begin{align*}\mu([r,1)) = \int_{r}^1 \mathrm{d} \mu(t)> 0.\end{align*}$$

If not, there exists $r_0 \in (0,1)$ such that $\mu ([r_0,1)) = 0$ . Then

$$\begin{align*}\int_{0}^{1} r^{n} \mathrm{d} \mu(r) = \int_{0}^{r_0}r^n \mathrm{d} \mu(r) \le r_0^n \mu([0,1)), \end{align*}$$

and note that

$$\begin{align*}\int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) = \sum_{k = 0}^{n} \int_0^1 r^k \mathrm{d} \nu(r) \le (n + 1) \nu([0,1]). \end{align*}$$

Hence,

$$\begin{align*}\int_0^1 \frac{1 - r^{n + 1}}{1 - r}\mathrm{d} \nu(r) \int_{0}^{1} r^n \mathrm{d} \mu(r) \le (n + 1)r_0^{n} \mu([0,1)) \nu([0,1]). \end{align*}$$

The right-hand side of the inequality converges to zero as $n \to \infty $ , which is a contradiction.

Hence,

$$\begin{align*}\omega([r,1)) = \frac{1}{2} \int_{[r,1)} t^{-1} \mathrm{d} (\psi_*\mu)(t) = \frac{1}{2} \int_{[r^2,1)} s^{-1/2} \mathrm{d} \mu(s) \geq \frac{1}{2} \mu([r^2,1))> 0. \end{align*}$$

It remains to show that $\omega $ is a $\widehat {\mathcal {D}}$ -weight. This is equivalent to showing [Reference Peláez, Rättyä and WickPRW19] that there exists a finite positive constant C such that

$$ \begin{align*}\omega_n \le C \omega_{2n}\end{align*} $$

for all $n \in \mathbb {N}$ , where $\omega _n = \int _0^1 r^n \mathrm {d} \omega (r)$ .

The case $n=0$ is trivial. Let $n\ge 1$ . By the definition (3.2) of the measure $\omega $ and a change of variables, we have

$$\begin{align*}\omega_n = \int_0^1 t^n \mathrm{d} \omega(t) =\frac{1}{2}\int_0^1 t^{n-1} \mathrm{d} (\psi_*\mu)(t) =\frac{1}{2}\int_0^1 t^{(n-1)/2} \mathrm{d} \mu(t). \end{align*}$$

Here $\psi (x) = \sqrt {x}$ . By Lemma 3.3, for every $s\ge 0$ ,

$$\begin{align*}\int_0^1 t^s \mathrm{d} \mu(t)=\frac{1}{f(s+1)}. \end{align*}$$

Therefore, for every $n\ge 1$ ,

$$\begin{align*}\omega_n=\frac{1}{2f \left(\frac{n+1}{2}\right)}, \qquad \omega_{2n}=\frac{1}{2f \left(n+\frac{1}{2}\right)}. \end{align*}$$

Thus

$$\begin{align*}\frac{\omega_n}{\omega_{2n}} = \frac{f \left(n+\frac{1}{2}\right)}{f \left(\frac{n+1}{2}\right)}. \end{align*}$$

It remains to prove that

$$\begin{align*}f \left(n+\frac{1}{2}\right)\le 2 f \left(\frac{n+1}{2}\right), \qquad n\ge 1. \end{align*}$$

For $k\ge 0$ , write $\nu _k:=\int _0^1 r^k \mathrm {d} \nu (r).$ Since $\nu $ is a finite positive measure on $[0,1]$ , the sequence $(\nu _k)_{k\ge 0}$ is decreasing, and f is increasing. We now distinguish two cases.

Case 1: $n=2m-1$ for $m\ge 1$ . Then

$$\begin{align*}f \left(n+\frac{1}{2}\right)= f(2m - \frac{1}{2}) \le f(2m)=\sum_{k=0}^{2m-1}\nu_k, \qquad f \left(\frac{n+1}{2}\right)=f(m)=\sum_{k=0}^{m-1}\nu_k. \end{align*}$$

Since $(\nu _k)$ is decreasing, we have

$$ \begin{align*}\sum_{k=m}^{2m-1}\nu_k \le \sum_{k=0}^{m-1}\nu_k.\end{align*} $$

Hence,

$$\begin{align*}f(2m)=\sum_{k=0}^{m-1}\nu_k+\sum_{k=m}^{2m-1}\nu_k \le 2\sum_{k=0}^{m-1}\nu_k =2f(m). \end{align*}$$

Therefore,

$$\begin{align*}f \left(n+\frac{1}{2}\right)\le 2 f \left(\frac{n+1}{2}\right). \end{align*}$$

Case 2: $n=2m$ for $m\ge 1$ . Since f is increasing, we get

$$\begin{align*}f \left(n+\frac{1}{2}\right)=f \left(2m+\frac{1}{2}\right)\le f(2m+1)=\sum_{k=0}^{2m}\nu_k. \end{align*}$$

On the other hand,

$$ \begin{align*} f \left(\frac{n+1}{2}\right) =f \left(m+\frac{1}{2}\right) &=\int_0^1 \frac{1-r^{m+1/2}}{1-r} \mathrm{d} \nu(r)\\ &=\sum_{k=0}^{m-1}\nu_k+\int_0^1 r^m\frac{1-r^{1/2}}{1-r} \mathrm{d} \nu(r). \end{align*} $$

Since, for every $r\in (0,1)$ ,

$$\begin{align*}\frac{1 - r^{1/2}}{1 - r} = \frac{1}{1 + r^{1/2}} \ge \frac{1}{2}, \end{align*}$$

we have

$$\begin{align*}f \left(\frac{n+1}{2}\right) \ge \sum_{k=0}^{m-1}\nu_k+\frac{1}{2}\int_0^1 r^m \mathrm{d} \nu(r) = \sum_{k=0}^{m-1}\nu_k + \frac{1}{2} \nu_m. \end{align*}$$

Moreover, using again that $(\nu _k)$ is decreasing, we have

$$\begin{align*}\sum_{k=m+1}^{2m}\nu_k \le \sum_{k=0}^{m-1}\nu_k. \end{align*}$$

Therefore,

$$ \begin{align*} f \left(n+\frac{1}{2}\right) &\le \sum_{k=0}^{2m}\nu_k =\sum_{k=0}^{m-1}\nu_k+\nu_m+\sum_{k=m+1}^{2m}\nu_k\\ &\le 2\sum_{k=0}^{m-1}\nu_k+\nu_m = 2\left(\sum_{k=0}^{m-1}\nu_k+\frac{1}{2} \nu_m\right)\\ &\le 2 f \left(m+\frac{1}{2}\right) =2 f \left(\frac{n+1}{2}\right). \end{align*} $$

Combining the two cases, we obtain

$$\begin{align*}f \left(n+\frac{1}{2}\right)\le 2 f \left(\frac{n+1}{2}\right), \qquad n\ge 1. \end{align*}$$

Hence, for every $n\ge 1$ ,

$$\begin{align*}\frac{\omega_n}{\omega_{2n}} = \frac{f \left(n+\frac{1}{2}\right)}{f \left(\frac{n+1}{2}\right)} \le 2. \end{align*}$$

Together with the trivial case $n=0$ , this shows that

$$\begin{align*}\omega_n\le 2 \omega_{2n}, \qquad n\in \mathbb N. \end{align*}$$

Therefore, $\omega $ is a $\widehat {\mathcal {D}}$ -weight. This completes the proof.