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Let $L>0$ and $\Omega \subset \mathbb{R}^n$ a bounded Lipschitz domain. Define $$ B_{\frac12,L}:=\{f \in L^2((0,1) \times \Omega) : \|f(t,\cdot)-f(s,\cdot)\|_{L^2(\Omega)} \leq L|t-s|^{\frac12},~ \forall s,t \in [0,1]\}. $$ I would like to show that, for every fixed $s,t \in [0,1]$, the functional $f \mapsto \int_\Omega |f(t,x)-f(s,x)|$ is continuous with respect to $L^2((0,1) \times \Omega)$-norm on the set $B_{\frac12,L}$, . It seems to me that this amounts to the question of continuity of point evaluation on a set of Hölder continuous functions with respect to $L^2$-norm, but I was not able to show it.

Does anyone have any direction: reference, counterexample, proof (hopefully)?

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  • $\begingroup$ I have troubles parsing what your actual question, or rather the actual functional, is. Do you want $f \mapsto \sup_{t,s \in [0,1]} \int_\Omega |f(t,x)-f(s,x)| \, \mathrm{d}x$ continuous with respect to the $L^2(I \times \Omega)$ norm on $B_{1/2,L}$? $\endgroup$
    – Hannes
    Aug 12, 2021 at 8:48
  • $\begingroup$ I edited the question to clarify a bit. So, for every fixed $s,t$ you want that $|Af-Ag|$ is small for small $\|f-g\|_{L^2(I\times \Omega)}$, where $A$ is the operator defined above. $\endgroup$ Aug 12, 2021 at 9:35

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I believe that I found a way to solve the problem. Let us show that the point evaluation $f \mapsto f_t$ is a bounded operator from $L^2((0,1) \times \Omega)$ to $L^2(\Omega)$ and the rest follows analogously. Namely, from the given condition, for every $s \in [0,1]$ we have $\|f_t\|_{L^2(\Omega)}^2 \leq 2(L^2|t-s| + \|f_s\|_{L^2(\Omega)}^2) $ and thus $ \sup_{\|f\|_{L^2((0,1)\times \Omega)}=1} \|f_t\|_{L^2(\Omega)} \leq \sqrt{2L^2+2} $, for every $t \in [0,1]$, from where the boundedness follows.

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