Let $f:[0,1]^2\rightarrow \mathbb{R}$ be a twice continuously differentiable function with the property that for all $x\in [0,1]$, there is an interval $I_x\subset [0,1]$ such that $f(x,y)=0$ for all $y\in I_x$. Does it follow that there must exist an open ball in $[0, 1]^2$ where $f$ is identically $0$?
11
$\begingroup$
$\endgroup$
Yes. Each $I_x$ contains some segment $[p,q]$ with rational endpoints. Let $A(p,q)$ be a set of corresponding $x$. It is impossible by Baire category theorem that each $A(p,q)$ is nowhere dense. Thus there exist $p,q$ and a segment $\Delta$ of positive length such that $A(p,q)$ is dense in $\Delta$. Hence $f$ does vanish on $\Delta\times [p,q]$, continuity of $f$ is enough here.
