Silly question on lower semi-continuity

Question

Suppose that $X_n\rightarrow X$ in a complete separable metric space $(\mathcal{X},d)$. Let $f:\mathcal{X}\rightarrow (-\infty,\infty]$ be a proper, convex, lower semi-continuous function, such that $$ f(X_{n+1})\geq f(X_n) (\forall n \in \mathbb{N}). $$ Can we conclude that $f(\lim_{n \mapsto0}X_n)\geq \limsup_{n \mapsto0}f(X_n)$?

It seems to me to be false, but I cant prove it, which isn't a good sign generally...

Answer 1

The answer is no. Let $$ \Phi\colon L^2(0,1)\to[0,\infty],\,\Phi(f)=\begin{cases}\int_0^1|f'|^2&\text{if }f\in W^{1,2}(0,1),\\\infty&\text{otherwise.}\end{cases} $$ Clearly $\Phi$ is a proper convex lower semicontinuous functional (the last property follows from the completeness of $W^{1,2}$).

The sequence $(f_n)$ given by $f_n(x)=x^n$ converges to $0$ in $L^2$, $\Phi(f_n)=n^2/(2n-2)$ is increasing in $n$, but $\Phi(0)=0<\infty=\limsup_n \Phi(f_n)$.

Answer 2

Let $f(x)\triangleq I(x>0)$, $X_n\triangleq \frac1{n},X\triangleq 0$.
Then $f(X_n)=1$ for every $n$; hence its $\limsup$ is $1$, which is bigger than $I(x>0)(X)=0$.