6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Jack M Jul 12 '18 at 9:02
  • $\begingroup$ @JackM Interesting, that would be theorem 1.2 in this paper: msp.org/pjm/1974/51-2/pjm-v51-n2-p23-s.pdf $\endgroup$ – punctured dusk Jul 12 '18 at 9:18
  • $\begingroup$ 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... $\endgroup$ – punctured dusk Jul 12 '18 at 9:18
  • $\begingroup$ 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. $\endgroup$ – Jack M Jul 12 '18 at 9:36
6
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ $x^3 + y^3$ 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. $\endgroup$ – Daniel Fischer Jul 12 '18 at 11:01
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.