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Let $f : \mathbb R \times \mathbb R \to \mathbb R$ be continuous when we fix one variable. Then $f$ need not be continuous (see e.g. Functions continuous in each variable ).

Does it imply that $f$ is locally bounded?

I would be surprised, but couldn't immediately think of a counterexample.

Context: I'm interested in $f : \mathbb R \times \mathbb C \to \mathbb C$ that are continuous in the first and analytic in the second variable. In this case, Cauchy's integral formula + Dominated convergence tells us that locally bounded implies jointly continuous.

Bart Michels
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    A function continuous in each variable is jointly continuous on a dense set of points, so it's at least locally bounded on a dense set of points. – Jack M Jul 12 '18 at 09:02
  • @JackM Interesting, that would be theorem 1.2 in this paper: https://msp.org/pjm/1974/51-2/pjm-v51-n2-p23-s.pdf – Bart Michels Jul 12 '18 at 09:18
  • Unfortunately, we can't guarantee that there are contours in this dense set, so the question about $\mathbb R \times \mathbb C$ remains a mystery... – Bart Michels Jul 12 '18 at 09:18
  • I believe it's a theorem of Baire, originally. If I can plug my own work for a minute, I posted a more elementary proof here. – Jack M Jul 12 '18 at 09:36

2 Answers2

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$f(x,y)=\frac {xy} {x^{3}+y^{3}}$ if $(x,y) \neq (0,0)$, $0$ if $(x,y) = (0,0)$. Note that $f(x,x)$ is not bounded near $0$.

Kavi Rama Murthy
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    $x^3 + y^3$ vanishes on the line $x + y = 0$. The obvious fixes are a) taking an even exponent ($> 2$) in the denominator or b) use $\lvert x\rvert^3 + \lvert y\rvert^3$. See which you prefer. – Daniel Fischer Jul 12 '18 at 11:01
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We can show that, for $K \subseteq \mathbb R$ compact and $U \subseteq \mathbb R$ open, there exists $V \subseteq U$ open such that $f : K \times V \to \mathbb R$ is bounded. That is, globally in one variable and locally in the other (but we have no control over $V$).

In particular, there is a dense open set $V \subseteq \mathbb R$ such that $f : K \times V \to \mathbb R$ is locally bounded.

It follows by applying the Baire category theorem to the closed sets $$\Omega_B = \{ y \in U : \forall x \in K : |f(x, y)| \leq B \}$$

We use continuity in the second variable to have that the $\Omega_B$ are closed, and in the first variable to have that their union is $U$. I found the argument at the bottom of page 2 here: http://www-users.math.umn.edu/~garrett/m/complex/hartogs.pdf

Bart Michels
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