Prove that $\Phi$ is a lower semicontinuous functional.

Question

QUESTION: Let $(X, \|.\|)_X$ be a Banach space and $Y\subset X$ a subspace, which is itself a Banach space endowed with a norm $\|.\|_Y$ such that $\|y\|_X\leq \|x\|_Y$ for every $y\in Y$. Assume that $J:X \rightarrow \mathbb{R}$ is a functional on $X$ such that $J|_{(Y, \|.\|)_Y}$ is continuous and satisfies:

1. $J$ has a directional derivative $\langle J'(u), v\rangle$ at each $u\in X$ through any direction $v\in Y$.
2. For fixed $u\in X$, the function $\langle J'(u), v\rangle$ is linear in $v\in Y$, and, for fixed $v\in Y$, the function $\langle J'(u), v \rangle$ is continuous in $u\in X$.

Let $K$ be a compact metric space, $K_0\subset K$ a closed subset and $\gamma_0: K_0\longrightarrow (Y, \|.\|_Y)$ a continuous map. Consider that $$\Gamma=\left\{\gamma:K\longrightarrow (Y, \| . \|) : \gamma\; \text{is continuous and} \; \gamma|_{K_0}=\gamma_0\right\}$$ is a complete metric space endowed with the uniform distance $$d_{\Gamma}(\gamma_1, \gamma_2)=\displaystyle\max_{t\in K} \|\gamma_1(t)-\gamma_2(t)\|_Y,$$ with $\gamma_1, \gamma_2\in \Gamma$. Defining, $\Phi$ on $\Gamma$ by setting $\Phi(\gamma)=\displaystyle\max_{t\in K} J(\gamma(t))$ prove that $\Phi$ is a lower semicontinuous functional.

MY ATTEMPT: I understood that $\Phi$ as defined above is like

\begin{alignat*}{2} \Phi:\Gamma &\longrightarrow X &\\ \gamma &\longmapsto & \Phi(\gamma):(Y, \Vert\cdot\Vert_{Y}) &\longrightarrow\mathbb{R} \\ &&J&\longmapsto\Phi(\gamma)=\displaystyle\max_{t\in K} J(\gamma(t)),\; \forall \gamma \in \Gamma \end{alignat*} Then I began to proof as below:

Assuredly, let $(\gamma_n)$ be a sequence of functions in $\Gamma$ such that $\gamma_n\rightarrow\gamma$ in $\Gamma$, what is true once $\Gamma$ is complete. Hence $\Phi(\gamma_n)\rightarrow\Phi(\gamma)$ in $\mathbb{R}$. Thus, $\Phi(\gamma_n)$ is bounded. One wants to prove that, for every $\lambda\in \mathbb{R}$ the set $$[\Phi\leq \lambda]:=\{\gamma\in \Gamma: \Phi(\gamma)\leq \lambda\}\; \text{is closed.}$$ Indeed, let $(\gamma_n)$ be a sequence of functions in $[\Phi\leq \lambda]$, then \begin{align*} \lim_{n\to \infty}\Phi(\gamma_n)&\leq \lim_{n\to \infty}\lambda=\lambda\\ \Phi(\gamma)=\Phi(\lim_{n \to \infty}\gamma_n)&\leq \lambda \end{align*} Therefore, $[\Phi\leq \lambda]$ is closed, which means that $\Phi$ is l.s.f.

DOUBT: Would someone tell me if my proof is right? Thanks in advance.

Answer 1

As already pointed out in the comment, your proof is wrong, because you don't know yet if $\Phi$ is continuous, hence you cannot pull limits in or out just now.

Instead consider the following hints:

1. Write $\Phi(\gamma)=\Phi(\gamma-\gamma_n)+\Phi(\gamma_n)$ and use a simple $\varepsilon$-limit argument.

2. We are only interested in $J$ restricted to $Y$, what can you say about a continuous linear map?

3. Deduce the result.