Consider $f:\mathbb{R}^2\to \mathbb{R}$ a function continuous in each variable separately. Suppose that the image of a compact subset of $\mathbb{R}^2$ is a compact subset of $\mathbb{R}$. Prove that $f$ is continuous.

I know that separately continuity does not imply continuity, it's enough consider $(x,y) \mapsto \frac{xy}{x^2+y^2}$ if $(x,y)\neq(0,0)$ and $(0,0)\mapsto 0$ as counterexample, however I don't know how to prove that $f$ with the additional hypothesis above is continuous.

  • $\begingroup$ Are you familiar with the "double subsequence trick"? $\endgroup$ – Nate Eldredge May 1 '18 at 16:20
  • $\begingroup$ No. What is it? $\endgroup$ – João Costa May 1 '18 at 22:04

Let $p_n\to p$ be a convergent sequence of different points on $\mathbb{R}^2$.

Then $K=\{p_n:\ n\in\mathbb{N}\}\cup\{p\}$ is compact. Therefore, its image is compact.

If $f(K)$ accumulates only at $f(p)$ then $f(p_n)\to p$.

Assume that $f(K)$ accumulates at $q\neq f(p)$.

If there is a subsequence $p_{m_n}$ of $p_n$ such that $f(p_{m_n})\neq q$ and $f(p_{m_n})\to q$, then dropping from $K$ all points at which $f(x)=q$, we get a new compact, which image is not compact.

Assume that we have a sequence $p_{m_n}$, such that $f(p_{m_n})=q$. Then, there must be some point $q_n=$ $p_{m_n}+t(0,1)$ or $p_{m_n}+t(1,0)$ with $|t|<a_n$ with $|f(q_n)-q|\to 0$. For some sequence $a_n\to0$.

If it is possible to procure infinitely many such $q_n$ with $f(q_n)\neq q$, then $\{q_n\}\cup\{p\}$ is compact, but its image is not compact.

If it is not possible to procure the sequence $q_n$, then each $p_{m_n}$ has a little cross with arms parallel to the axes on which $f$ is constant and equal to $q$. Moreover, the length of the arms do not tend to zero. But then the arms of those crosses intersect the lines parallel to the axes passing through $p$, at arbitrarily close distance from $p$. But that contradicts that $f$ is continuous at $p$, separately on each variable.

| cite | improve this answer | |

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.