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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.

Lemon
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    What is $H^1(\Omega)$? Forgive my lack of knowledge of functional analysis. – akkkk Dec 22 '12 at 16:53
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    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
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    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
  • @akkkk $H^1$ is the Sobolev space with 1 weak derivative. – Lemon Dec 22 '12 at 18:56
  • @GEdgar But the supremum is taken wrt. $u$ here. I only want continuity of $t$ – Lemon Dec 22 '12 at 18:57
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    @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
  • ... and since the constant, integer-valued functions are discrete in $H^1(\Omega)$, @GiuseppeNegro's example can be interpolated/extrapolated in many ways to be a continuous function on $H^1$... – paul garrett Dec 22 '12 at 21:14
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    @GiuseppeNegro Thanks – Lemon Dec 23 '12 at 13:17
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    @paulgarrett thanks – Lemon Dec 23 '12 at 13:18

1 Answers1

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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.

Greg Martin
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  • Thanks but your $f$ is not continuous at $t=1$ I believe. – Lemon Dec 22 '12 at 18:58
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    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 01:21