Let $X$ be a Banach sapce and $f$ be a lower semicontinuous mapping from $X$ to $\mathbb{R}$. We consider the following definitions:
$x_0\in X$ is called a strict minimizer of $f$ if $f(x)>f(x_0)$ for all $x\in X$ and $x\ne x_0$;
$x_0\in X$ is called a strong minimizer of $f$ if $f(x_0)=\inf_Xf$ and every sequence $(x_n)$ along which $f(x_n)\rightarrow f(x_0)$ obeys $\|x_n-x_0\|\rightarrow 0$;
$x_0\in X$ is called a strict local minimizer of $f$ if there exists $\varepsilon>0$ such that $f(x)>f(x_0)$ for all $x\in X$ such that $x\ne x_0$ and $\|x-x_0\|\leq\varepsilon$;
$x_0\in X$ is called a strong local minimizer of $f$ if there exists $\varepsilon>0$ such that $f(x_0)=\inf_{\|x-x_0\|\leq\varepsilon}f(x)$ and every sequence $(x_n)$ along which $\|x_n-x_0\|\leq\varepsilon$ and $f(x_n)\rightarrow f(x_0)$ obeys $\|x_n-x_0\|\rightarrow 0$;
From definitions, it is easy to verify that:
strong minimizer is strict minimizer
strong local minimizer is strict local minimizer
For the function $f(x)=x^2e^x$, $x_0=0$ is the strict minimizer of $f$ but $x_0$ is not the strong minimizer of $f$.
Question. I would like to construct a lower semicontinuous function $f:X\rightarrow\mathbb{R}$ and a point $x_0$ such that $x_0$ is the strict local minimizer of $f$ but $x_0$ is not the strong local minimizer of $f$.
My attempt. When $X=\mathbb{R}^n$ and $f$ is lower semicontinuous, we could prove that strong local minimizer is actually strict local minimizer. Therefore, we have to to construct counterexample in infinite dimensional setting.
Thank you for all kind help and suggestion.
Take $X=L^2(0,1)$, $f(x)=\|x\|_{L^1}$, $x_0=0$. Define $$ x_n = n \chi_{[0,n^{-2}]} . $$ Hence $\|x_n\|_{L^2}=1$ and $f(x_n)=\|x_n\|_{L^1}=n^{-1}\to 0$, but $x_n$ does not converge to $0$ in $L^2$. This shows that $x_0=0$ is strict local but not strong local minimizer.
Here is another example, where the sequence $(x_n)$ does not even converge weakly. Take $X=L^1(\mathbb R)$, take $a(t)=\frac1{1+t^2}$, set $f(x)=\int_{\mathbb R} a|x| dt$. Then $x_0=0$ is the unique minimum. Define $x_n = \chi_{[n,n+1]}$. Then $(x_n)$ is bounded, $f(x_n)$ converges to zero. But $(x_n)$ is not converging weakly to zero as $\int_{\mathbb R}x_n=1$.