On the zero set of a $C^2$ function on $[0,1]^2$

Question

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 $0$?

Answer 1

Let $I_n$ be an enumeration of the intervals $(a,b)$ where $0\le a < b\le 1$ and $a,b \in \mathbb {Q}.$ Let $E_n = \{x\in [0,1]: f(x,y)=0\ \text {for}\ y \in I_n\}.$ The given property implies $[0,1] = \cup E_n.$ The continuity of $f$ shows each $E_n$ is a closed subset of $[0,1].$ Thus by Baire, some $E_n$ contains an open interval $J.$ It follows that $f = 0$ on $J\times I_n,$ which certainly contains an open ball. (Only the continuity of $f$ was used here.)