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Given a function $f(x,y)\in L^p(R^d;L^\infty(B_R))$ with $1<p<\infty$, where $B_R:=\{y\in R^d: |y|\le R\}$, can we find a sequence of functions $f_n$ of the form $f_n(x,y)=\sum_{i=1}^ng_i(x)h_i(y)$ such that $$ \big\|\sup_{|y|\le R}|f_n(\cdot,y)-f(\cdot,y)|\big\|_{L^p}\to 0\quad as\quad n\to\infty $$ and $$ |f_n(x,y)|\leq |f(x,y)|??? $$

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  • $\begingroup$ Could you add some details (just for clarity): what do you mean precisely by $f(x,y)\in L^p(R^d;L^\infty(B_R))$? $\endgroup$ – Skeeve Mar 27 '19 at 7:55
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I think this is not possible.

Call $f(x,y)$ the characteristic function of the set $\{x<y\}$ in $R^2$. If $h(y)$ is any function of the single variable $y$, we have $\sup_y|f(x,y)-h(y)|\ge 1/2$ for a.e. $x$. Now, suppose you can approximate $f$ (locally) with tensor products; then you can approximate $f$ with tensor products of simple functions. Let $u(x,y)=\sum g_i(x)h_i(y)$ be any such function. Represent all $g_i$ in the form $g_i(x)=\sum_{k=1}^N c_{ik}\chi_{E_k}$ with the same $N$ and the same sets $E_k$. You see that for $x\in E_1$ we have $u(x,y)=\sum c_{1k}h_k(y)$ independent of $x$, hence $\sup_y|f(x,y)-u(x,y)|\ge 1/2$ on $E_1$. The same argument applies to all $E_i$ and we have a contradiction.

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  • $\begingroup$ Maybe there is a small typo in the formula for $u(x,y)$ (no $y$ in the right hand side). And what if $g_i$ are not simple? $\endgroup$ – Skeeve Mar 27 '19 at 7:57
  • $\begingroup$ You can approximate a tensor product by a tensor product of simple functions. I corrected the typo, thanks $\endgroup$ – Piero D'Ancona Mar 27 '19 at 8:01
  • $\begingroup$ One more question: your argument implies that $f$ is not strongly measurable (in the Bochner sense), hence $f\notin L^p(\mathbb R; L^\infty(\mathbb R))$, right? $\endgroup$ – Skeeve Mar 28 '19 at 8:05
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    $\begingroup$ Right, if you regard that space as a Bochner space than $f$ does not belong to it. However in applications to PDEs the $L^pL^q$ spaces are usually defined (implicitly) as spaces of Lebesgue measurable functions in $x,y$ for which the $L^pL^q$ norm is finite $\endgroup$ – Piero D'Ancona Mar 28 '19 at 8:11

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