Approximation of functions by tensor products

Given a function $$f(x,y)\in L^p(R^d;L^\infty(B_R))$$ with $$1, where $$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)|???$$

• Could you add some details (just for clarity): what do you mean precisely by $f(x,y)\in L^p(R^d;L^\infty(B_R))$? Mar 27 '19 at 7:55

Call $$f(x,y)$$ the characteristic function of the set $$\{x in $$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.
• 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? Mar 27 '19 at 7:57
• 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? Mar 28 '19 at 8:05
• 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 Mar 28 '19 at 8:11