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The Art of Mathematics: Coffee Time in Memphis, page 77
22. Separately Continuous Functions. Let $f: S=[0,1]^{2} \rightarrow \mathbb{R}$ be separately continuous in its variables, i.e., continuous in $x$ for every fixed $y$, and continuous in $y$ for every fixed $x$. Show that if $f^{-1}(0)$ is dense in the square $S$ then it is identically 0 .
Proof. Suppose for a contradiction that $f$ is not identically 0 . Then we may assume that $f\left(x_{0}, y_{0}\right) \geq 3$, where $0<x_{0}<1$ and $0<y_{0}<1$.
Since $f\left(x, y_{0}\right)$ is a continuous function of $x$, there is an interval $I=[a, b]$, $0 \leq a<b \leq 1$, such that $f\left(x, y_{0}\right) \geq 2$ for $x \in I$.
Now, for $n \geq \min \left\{y_{0}, 1-y_{0}\right\}^{-1}$, set
$$
F_{n}=\left\{x \in I: f\left(x, y_{0}+y\right) \geq 1 \text { if }|y| \leq 1 / n\right\} .
$$
As $f$ is continuous in $x$, each $F_{n}$ is a closed subset of $I$. Furthermore, since $f$ is continuous in $y$ and $f\left(x, y_{0}\right) \geq 2>1$ for every $x \in I$, we have $I=\bigcup_{n} F_{n}$. By the Baire category theorem (see, e.g., p. 76 of B. Bollobás, Linear Analysis, Cambridge University Press, 1990 , xi +240 pp.), some $F_{n}$ has a non-empty interior, i.e., contains an interval $I_{n}=[c, d]$ with $0 \leq c<d \leq 1$. Consequently, $f(x, y) \geq$ 1 on the rectangle $(x, y) \in I_{n} \times\left[y_{0}-1 / n, y_{0}+1 / n\right] \subset S$, contradicting the assumption that $f^{-1}(0)$ is dense in $S$.
Notes. This was one of my favourite exercises for the Linear Analysis course in Cambridge in the early 1970s. |
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