Claim $\limsup _{t\to 0}[\max _{u\in S_{t_{0}+t(1-d_{t})}}\int _{\Omega }f^{\ast }(u)u\,{\textrm {d}}x]\leq \max _{u\in S_{t_{0}}}\int _{\Omega }f^{\ast }(u)u\,{\textrm {d}}x$.

Let $\{t_{n}\}$ be a sequence of real numbers converging to zero. For each $n$, choose $u_{n}\in S_{t_{0}+t_{n}(1-d_{t_{n}})}$ such that

\[ \int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x=\max_{u\in S_{t_{0}+t_{n}(1-d_{t_{n}})}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x. \]

Since $\|u_{n}\|^{2}\leq t_{0}+t_{n}(1-d_{t_{n}}),$ we have that $\{u_{n}\}$ is bounded in $H_{0}^{1}(\Omega )$ and, as before, there exists $u_{0}\in H_{0}^{1}(\Omega )$ such that, up to a subsequence,

\[ u_{n}\rightharpoonup u_{0}\ \mbox{in} H_{0}^{1}(\Omega), \]

and

\[ \|u_{0}\|^{2}\leq \liminf_{n\to \infty}\|u_{n}\|^{2}\leq t_{0}. \]

Moreover, by the continuity of $\alpha$,

\[ \alpha(t_{0})=\lim_{n\to\infty}\alpha(t_{0}+t_{n}(1-d_{t_{n}}))=\lim_{n\to\infty}\int_{\Omega}F^{{\ast}}(u_{n})\,{\textrm{d}}x=\int_{\Omega}F^{{\ast}}(u_{0})\,{\textrm{d}}x. \]

Consequently, $u_{0}\in S_{t_{0}}$ and, by $(i)$, $\|u_{0}\|^{2}=t_{0}$. The claim follows now by observing that the weak convergence of $u_n$ to $u_0$ implies that

\begin{align*} \limsup_{n\to \infty} \max_{u\in S_{t_{0}+t_{n}(1-d_{t_{n}})}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x & = \limsup_{n\to \infty}\int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x\\ & =\int_{\Omega}f^{{\ast}}(u_{0})u_{0}\,{\textrm{d}}x\\ & \leq \max_{u\in S_{t_{0}}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x. \end{align*}

Now, by previous claim, passing to the upper limit as $t$ tends to zero in (2.5), we get

\[ \limsup_{t\to 0}\left[\frac{\alpha(t_{0}+t)-\alpha(t_{0})}{t}\right]\leq \frac{1}{2t_{0}}\max_{u\in S_{t_{0}}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x. \]

By comparing this inequality with (2.4), we conclude that $\alpha$ is differentiable in $t_{0}$ and

\[ \alpha'(t_{0})=\frac{1}{2t_{0}}\max_{u\in S_{t_{0}}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x. \]

To prove the upper semicontinuity of the derivative, let $\{t_{n}\}$ be a sequence of real numbers converging to a positive number $t_{0}$. For each $n$, choose $u_{n}\in S_{t_{n}}$ such that

\[ \int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x=\max_{u\in S_{t_{n}}}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x. \]

Since $u_{n}\in S_{t_{n}}$, as previously, it follows that there exists $u_{0}\in H_{0}^{1}(\Omega )$ such that, up to a subsequence,

\begin{align*} & u_{n}\rightharpoonup u_{0}\ \mbox{in} H_{0}^{1}(\Omega),\\ & \|u_{0}\|^{2}\leq \liminf_{n\to \infty}\|u_{n}\|^{2}\leq t_{0},\\ & \lim_{n\to\infty}\alpha(t_n) = \lim_{n\to\infty} \int_{\Omega}F^{{\ast}}(u_{n})\,{\textrm{d}}x = \int_{\Omega}F^{{\ast}}(u_{0})\,{\textrm{d}}x,\\ & \lim_{n\to\infty} \int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x= \int_{\Omega}f^{{\ast}}(u_{0})u_{0}\,{\textrm{d}}x. \end{align*}

Using the continuity of $\alpha$,

\[ \alpha(t_{0})=\int_{\Omega}F^{{\ast}}(u_{0})\,{\textrm{d}}x, \]

and consequently

\[ u_{0}\in S_{t_{0}}, \]

and by the computation of the derivative $\alpha '(t_0)$, we obtain

\[ \limsup_{n\to\infty}\alpha'(t_{n})=\limsup_{n\to\infty}\frac{1}{2t_{n}}\int_{\Omega}f^{{\ast}}(u_{n})u_{n} \,{\textrm{d}}x=\frac{1}{2t_{0}}\int_{\Omega}f^{{\ast}}(u_{0})u_{0}\,{\textrm{d}}x\leq \alpha'(t_{0}). \]

To prove the last part of $(ii)$, let $t_{n}\searrow 0$ and $u_{n}\in S_{t_{n}}$ be such that

\[ \alpha'(t_{n})=\frac{1}{2t_{n}}\int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x. \]

Define $v_{n}:=u_{n}/t_{n}^{1/2}$ and observe that, by (i), $\|u_{n}\|=t_{n}^{1/2}$ and thus $\{v_{n}\}$ is bounded in $H_{0}^{1}(\Omega )$. Consequently, there exists $v_{\ast }$ such that, up to a subsequence,

(2.6)\begin{equation} v_{n}\rightharpoonup v_{{\ast}}\quad \mbox{in}\ H_{0}^{1}(\Omega). \end{equation}

Thus, by Lebesgue Dominated Convergence Theorem and the definition of $\gamma$, we get

(2.7)\begin{equation} \alpha'(t_{n})=\frac{1}{2t_{n}}\int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x=\frac{1}{2} \int_{\Omega}\left[\frac{f^{{\ast}}(t_{n}^{1/2}v_{n})}{t_{n}^{1/2}v_{n}}\right]v_{n}^{2}\,{\textrm{d}}x\to \frac{\gamma}{2}\int_{\Omega}v_{{\ast}}^{2}\,{\textrm{d}}x. \end{equation}

Moreover, by the definition of $\gamma$ again and the L'Hospital rule, we have $\lim _{t\searrow 0}F^{\ast }(t)/t^{2}=\gamma /2$. Thence, from (2.6), up to a subsequence, we obtain

\[ \lim_{n\to\infty} \frac{F^{{\ast}}(u_{n})}{t_{n}}=\lim_{n\to\infty}\left[\frac{F^{{\ast}}(t_{n}^{1/2}v_{n})}{(t_{n}^{1/2}v_{n})^{2}}\right]v_{n}^{2}=\frac{\gamma}{2} v_{{\ast}}^{2} \ \mbox{a.e. in}\ \Omega. \]

and

\[ \left| \frac{F^{{\ast}}(u_{n})}{t_{n}}\right|=\left|\frac{F^{{\ast}}(t_{n}^{1/2}v_{n})}{(t_{n}^{1/2}v_{n})^{2}}\right||v_{n}^{2}|\leq \gamma h(x) \ \mbox{a.e. in}\ \Omega, \]

for some $h\in L^{1}(\Omega )$. (Observe that the convergence (2.6) implies, passing to a subsequence if necessary, the existence of such a function $h(x)$). Therefore, by Lebesgue Dominated Convergence Theorem

(2.8)\begin{equation} \int_{\Omega}\frac{F^{{\ast}}(u_n)}{t_{n}}\,{\textrm{d}}x\to \frac{\gamma}{2}\int_{\Omega}v_{{\ast}}^{2}\,{\textrm{d}}x. \end{equation}

If it were $v_{\ast }=0$, then by (2.7) and (2.8), we would conclude that

(2.9)\begin{equation} 0=\lim_{n\to\infty}\alpha'(t_{n})=\frac{1}{2}\lim_{n\to\infty} \frac{1}{t_{n}}\int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x=\lim_{n\to\infty}\int_{\Omega}\frac{F^{{\ast}}(u_n)}{t_{n}}\,{\textrm{d}}x. \end{equation}

On the other hand, since $u_{n}\in S_{t_{n}}$, it follows again from Lebesgue Dominated Convergence Theorem and L'Hospital rule, that

(2.10)\begin{equation} \lim_{n\to\infty}\int_{\Omega}\frac{F^{{\ast}}(u_n)}{t_{n}}\,{\textrm{d}}x\geq \lim_{n\to\infty}\int_{\Omega} \frac{F^{{\ast}}(t_{n}^{1/2}\varphi_1)}{(t_{n}^{1/2}\varphi_1)^{2}}\varphi^{2}_{1}\,{\textrm{d}}x=\frac{\gamma}{2\lambda_1}, \end{equation}

where $\varphi _{1}$ is the eigenfunction of Laplacian operator with homogeneous Dirichlet boundary condition, associated with $\lambda _1$ with $\|\varphi _{1}\|=1$. By comparing (2.9) and (2.10), we get a contradiction proving that $v_{\ast }\neq 0$ $(v_{\ast }\geq 0)$.

Finally, since $u_{n}\in S_{t_{n}}$ and $\|u_{n}\|=t_{n}^{1/2}$, we can again use the Lagrange multipliers theorem (see proof of item $(i)$) to conclude that

\[ \left \{ \begin{array}{@{}ll} -\Delta u_n = \displaystyle\dfrac{1}{2\alpha'(t_{n})} f^{{\ast}}(u_n) & \mbox{in}\ \Omega,\\ u_n=0 & \mbox{on}\ \partial\Omega. \end{array}\right. \]

Consequently,

(2.11)\begin{equation} \int_{\Omega}\nabla v_{n}\nabla w \,{\textrm{d}}x=\displaystyle\frac{1}{2\alpha'(t_{n})} \int_{\Omega}\left[\frac{f^{{\ast}}(t_{n}^{1/2}v_{n})}{t_{n}^{1/2}v_{n}}\right]v_{n} w \,{\textrm{d}}x,\quad \forall\, w\in H_{0}^{1}(\Omega). \end{equation}

By passing to the limit as $n$ goes to infinity in (2.11), it follows from (2.6) and (2.7) that

\[ \int_{\Omega}\nabla v_{{\ast}}\nabla w \,{\textrm{d}}x= \frac{1}{\int_{\Omega}v_{{\ast}}^{2}\,{\textrm{d}}x} \int_{\Omega}v_{{\ast}} w \,{\textrm{d}}x, \quad \forall\, w\in H_{0}^{1}(\Omega). \]

Since $v^{*}\geq 0$, $v^{*}\not \equiv 0$, we have $\int _{\Omega }v_{\ast }^{2}\,{\textrm {d}}x=1/\lambda _{1}$ and (2.7) implies that $\lim _{n\to \infty } \alpha ' (t_n) = \gamma /(2\lambda _1)$.

Finally, suppose by contradiction that there exists a sequence of positive numbers $\{t_{n}\}$ and $\{u_{n}\} \subset S_{t_n}$ such that $t_{n}\to t_{\ast }$ (for some $t_{\ast }>0$), $\|u_{n}\|= t_{n}^{1/2}$, $\int _{\Omega }F^{\ast }(u_{n})\,{\textrm {d}}x=\alpha (t_{n})$ and

(2.12)\begin{equation} \alpha' (t_n)=\frac{1}{2t_{n}}\int_{\Omega}f^{{\ast}}(u_{n})u_{n}\,{\textrm{d}}x\to 0. \end{equation}

In particular, $\{u_{n}\}$ is bounded and hence there exists $u_{\ast }\in H_{0}^{1}(\Omega )$ with $u_{\ast }\geq 0$ (because $u_{n}>0$) and such that, up to a subsequence,

\[ u_{n}\rightharpoonup u_{{\ast}}\ \mbox{in} H_{0}^{1}(\Omega). \]

The weak continuity of $J$ and of the map $\int _{\Omega }f^{\ast }(v)v\,{\textrm {d}}x$, (2.12), $(i)$ and the continuity of $\alpha$ imply by taking limits when $n$ tends to infinity that

(2.13)\begin{equation} \int_{\Omega}F^{{\ast}}(u_{{\ast}})\,{\textrm{d}}x=\alpha (t_{{\ast}})>0 \end{equation}

and

(2.14)\begin{equation} \frac{1}{2t_{{\ast}}} \int_{\Omega}f^{{\ast}}(u_{{\ast}})u_{{\ast}}\,{\textrm{d}}x=0 \end{equation}

Identity (2.14) implies that $u_{\ast }=0$, but that leads us to a contradiction with (2.13). The result is now proved.

(iii) Let us consider first $k=2,\ldots , K$. By (ii), $g_{k}: (0, \infty )\to {\rm I}\hskip -0.85mm{\rm R}$ is differentiable with $g'_k (t) =m_k (t) - \alpha '(t)$ for every $t>0$. If $t\in (0, t_{k-1}]$, by the definition of $m_k$ we have $g_{k}(t)=-\alpha (t)$ and thus

\[ g_{k}'(t)={-}\alpha'(t)={-}\frac{1}{2t}\max_{u\in S_t}\int_{\Omega}f^{{\ast}}(u)u\,{\textrm{d}}x<0,\quad \forall \, t\in (0, t_{k-1}]. \]

On the other hand, by (m), $\lim _{t\searrow t_{k-1}} m_k(t)= m_k(t_{k-1})=0$ and $(ii)$

\[ \limsup_{t\searrow t_{k-1}}g_{k}'(t)=m_k(t_{k-1})-\liminf_{t\searrow t_{k-1}}\alpha'(t)={-}\liminf_{t\searrow t_{k-1}}\alpha'(t)<0. \]

Finally, assuming hypothesis (mf), if $k=1$ then, by $g_{1}(0)=0$ and $(ii)$, we have

\[ \lim_{t\searrow 0}g_{1}'(t)=\frac{1}{2}\left[m(0)-\frac{\gamma}{\lambda_1}\right]<0. \]

The result follows.