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 $$ f_n(x,y)\leq f(x,y)??? $$

$\begingroup$ Could you add some details (just for clarity): what do you mean precisely by $f(x,y)\in L^p(R^d;L^\infty(B_R))$? $\endgroup$– SkeeveMar 27 '19 at 7:55
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 $\sup_yf(x,y)h(y)\ge 1/2$ 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 $\sup_yf(x,y)u(x,y)\ge 1/2$ on $E_1$. The same argument applies to all $E_i$ and we have a contradiction.

$\begingroup$ 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? $\endgroup$– SkeeveMar 27 '19 at 7:57

$\begingroup$ You can approximate a tensor product by a tensor product of simple functions. I corrected the typo, thanks $\endgroup$ 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? $\endgroup$– SkeeveMar 28 '19 at 8:05

1$\begingroup$ 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 $\endgroup$ Mar 28 '19 at 8:11