A multiplier to make a couple of bounded functions continuous wrt one variable?

Question

Let $Q=(0,1)^2$ be a square, functions $f_1,f_2\in L_\infty(Q)$ and $f_1,f_2\ge c>0$ on $Q$. Does there exists for every pair of such $f_1,f_2$ a function $g>0$ on $Q$ s.t. function $f_1g$ is continuous wrt $x$ on $(0,1)$ for every $y\in(0,1)$ and $f_2g$ is continuous wrt $y$ for every $x\in(0,1)$?

Answer 1

If your question assumes that $f_1$ and $f_2$ are defined everywhere (and not almost everywhere as it is standard for $L^\infty$ functions)

Let $f_1$ be $2$ on a vortex (for exemple the image of $\frac{1+i}{2}+\frac{1}{t}e^{it}$) and 1 evrywhere else, and let $f_2$ be $1/f_1$.

To have the horizontal continuity of $f_1g$ we must have $g=C/f_1$ where $C$ is any continuous bounded away from zero function. But then $f_2 g=Cf_2^2$ is discontinuous.

If you wanted to know if we could find a function $g$ and a pair of representant of $f_1$ and $f_2$ such that you have these continuity properties, my counterexample needs some adaptation.

We consider a thickening of our vortex: $\mathcal{V}=\left\lbrace x\in Q | \exists t\in \mathbb ]0,+\infty[,\exists \theta \in [0, \pi/6], x=\frac{1+i}{2}+\frac{1}{t}e^{i(t+\theta)} \right\rbrace$

Suppose that our property is true, let $f_1$ and $f_2$ be representatives of $1+\chi_\mathcal{V}$ and $1-\frac{1}{2}\chi_\mathcal{V}$. Let $g>0$ satisfy our continuity property. We must have $g=C/f_1$ and $g=\widetilde{C}/f_2$. Impossible.

Hence the falsehood of our assumption.