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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.

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    What is $H^1(\Omega)$? Forgive my lack of knowledge of functional analysis.2012-12-22
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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 $H^1(\Omega)$.2012-12-22
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    Well, it has *almost* nothing to do with $H^1(\Omega)$. We need that $H^1(\Omega)$ isn't compact.2012-12-22
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    @akkkk $H^1$ is the Sobolev space with 1 weak derivative.2012-12-22
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    @GEdgar But the supremum is taken wrt. $u$ here. I only want continuity of $t$2012-12-22
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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 $x$.2012-12-22
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    ... 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 $H^1$...2012-12-22
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    @GiuseppeNegro Thanks2012-12-23
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    @paulgarrett thanks2012-12-23

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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.

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    Thanks but your $f$ is not continuous at $t=1$ I believe.2012-12-22
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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.2012-12-23