# Silly question on lower semi-continuity

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...

• You are using $X$ for both the space and an element in it!. – Kavi Rama Murthy Feb 6 at 10:11
• Thanks for pointing out the morning madness Ravi – AIM_BLB Feb 6 at 10:18
• What does it mean for $X$ to be convex on a general metric space $\mathcal X$? – Julián Aguirre Feb 6 at 10:21
• The space should be a Length space and then tranalate the usual geodesically conev definition...namesly the sum of the functional on the endpoint of any distance minimizing curve at least as much as its value on any single point in the distance minimizing curve – AIM_BLB Feb 6 at 10:23

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 $$\Phi(0)=0<\infty=\limsup_n \Phi(f_n)$$.
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 $$I(x>0)(X)=0$$.