Supremum of continuous functions is continuous?

If $f(t,u)$ is continuous wrt. $t$ (and $u$), then is $$\sup_{u \in H^1(\Omega)} f(t,u)$$ continuous wrt. $t$?

I am unable to prove this. Help appreciated.

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What is $H^1(\Omega)$? Forgive my lack of knowledge of functional analysis. –  akkkk Dec 22 '12 at 16:53
Generally, all we know is that the supremum of continuous functions is lower semicontinuous. Of course, that has nothing to do with $H^1(\Omega)$. –  GEdgar Dec 22 '12 at 17:00
Well, it has almost nothing to do with $H^1(\Omega)$. We need that $H^1(\Omega)$ isn't compact. –  Andreas Blass Dec 22 '12 at 17:14
@Lemon: I agree with GEdgar. Take the sequence $$f_n(x)= \begin{cases} 0 & x\le 0 \\ nx & 0< x < \frac{1}{n} \\ 1 & \frac{1}{n} \le x \end{cases},$$ where $n \in \mathbb{N}$ and $x\in \mathbb{R}$. If you let $$f(x)=\sup_{n \in \mathbb{N}}f_n(x)$$ you get the discontinuous function $$f(x)=\begin{cases} 1 & x \le 0 \\ 0 & 0< x\end{cases}$$ Here you took the supremum with respect to $n$ and obtained a discontinuous function of $x$. –  Giuseppe Negro Dec 22 '12 at 19:07
@GiuseppeNegro Thanks –  Lemon Dec 23 '12 at 13:17

Let $S= (0,\infty) \times (0,\infty)$. If $f\colon S\to\mathbb R$ is defined by $$f(t,u) = \begin{cases} 0, & \text{if } t\le 1, \\ (t-1)u, & \text{if } t>1 \text{ and } (t-1)u\le 1, \\ 1, & \text{if } t>1 \text{ and } (t-1)u>1, \end{cases}$$ then $$\sup_{u\in (0,\infty)} f(t,u) = \begin{cases} 0, & \text{if } t\le 1, \\ 1, & \text{if } t>1 \end{cases}$$ is not continuous.

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Thanks but your $f$ is not continuous at $t=1$ I believe. –  Lemon Dec 22 '12 at 18:58
I think it is continuous, actually. The third case of the definition doesn't come in to play infinitesimally to the right of $t=1$; it's always the second case of the definition that ends up being relevant. –  Greg Martin Dec 23 '12 at 1:21
Shouldn't $f(x)=\underset{n\in \mathbb{N}}{\mathop{\sup }}\,{{f}_{n}}(x)=\left\{ \begin{matrix} \color{red}0,\text{ if }x\le 0 \\ \color{red}1,\text{ if }x>0 \\ \end{matrix} \right.$ ?