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Consider $f\in L^2(\mathbb{R}^3)$ such that, denoting by $A_f$ the solution to the wave equation $\square A=0$ with initial data $A(0)=0$, $(\partial_t A)(0)=f$, we have $A\in L_2(\mathbb{R}^+;L^{\infty}(\mathbb{R}^3))$. Since it is known that the endpoint Strichartz estimates for the wave equation fails in dimension $d=3$, not every $f\in L^2$ would satisfy such condition. I would like to approximate (in the $L^2$-norm) $f$ with sufficiently smooth (say $H^2$) functions $f_n$, in such a way $\|A_{f_n}\|_{L_t^2L_x^{\infty}}$ is uniformly bounded in $n$. Is such approximation always possible? Thank you for any suggestions.

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    $\begingroup$ Just to clarify: you know $f\in L^2$ such that $A_f \in L^2L^\infty$, and you want approximations $f_n\in H^2$, converging to $f$ in $L^2$, such that $A_n \in L^2L^\infty$ is uniformly bounded? $\endgroup$ Commented Oct 21, 2020 at 16:40
  • $\begingroup$ Yes Willie, exactly. $\endgroup$
    – Capublanca
    Commented Oct 21, 2020 at 16:47
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    $\begingroup$ Couldn't you just mollify? Convolution with a spatial approximation to identity commutes with solving the wave equation (it is linear). And convolution decreases the $L^\infty$ norm. $\endgroup$ Commented Oct 21, 2020 at 17:37
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    $\begingroup$ I think you are right, thank you for the comment. I had some doubts cause if $f_{n} $ are the mollifications we don't have that they converge to $f$ in $L^{\infty}$, but for the uniform boundedness they seem to be sufficient. $\endgroup$
    – Capublanca
    Commented Oct 21, 2020 at 17:51

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