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Let $f \in H^{-1}(S)$, where $S \subset \mathbb{R}^n$ is some nice set in space. Can I exchange supremums and integrals over time here: $\sup_{g \in H^1(S)}\int_0^T \langle f, g \rangle_{H^{-1}(S), H^1(S)}\;dt =\int_0^T \sup_{g \in H^1(S)} \langle f, g \rangle_{H^{-1}(S), H^1(S)}\;dt$

I think so. But what if $g$ has some dependence on $t$? Or is that not allowed?

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    @HansEngler $S = \Omega$, some nice set. No time dependence.2012-12-12

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If $f$ depends on $t$ and supremum on the right is computed for each $t$, the two sides are not equal, even if $g$ does not depend on $t$.

A simple counterexample can be constructed by taking $T = 2 \pi, f(t) = \sin t \phi$, where $\phi$ is an arbitrary element in $H^{-1}(\Omega)$. Then if $g \in H^1(\Omega)$, the left hand side is zero, but the right hand side is positive.

If $g$ is allowed to depend on $t$, equality holds, I think.

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    @maximumtag Since $t$ has no meaning outside of the integral with respect to $t$, it's impossible for $g$ to depend on $t$ on the left side of the formula.2012-12-23