3
$\begingroup$

Question: Let $f:\mathbb{R^2} \to \mathbb{R}$ be a function such that:

(I) For all $y_0 \in \mathbb{R}$, $f(x,y_0)$ is continuous.

(II) For all $x_0 \in \mathbb{R}$, $f(x_0, y)$ is continuous.

(III) If $K\subset\mathbb{R^2}$ is compact, then $f(K)$ is compact.

Show that $f$ is continuous.

My attempt was: We shall prove that $\lim f(z_n) = f(\lim z_n)$. Take $(x_n,y_n) \in \mathbb{R^2}$ such that $(x_n,y_n) \to (x,y)$. Since $(x_n,y_n)$ is bounded, exists $K \subset \mathbb{R^2}$ compact such that $(x_n,y_n) \subset K$ and, by (III), $f(x_n,y_n)$ is bounded. By Bolzano-Weierstrass exists a subsequence convergent. Take any convergent subsequence $f(x_{n_k}, y_{n_k}) \to f(x',y')$. Note that $f(x',y') = \lim f(x_{n_k}, y')$, and, by (I), $\lim f(x_{n_k}, y') = f(x,y')$. Same argument using (II) for $y'$. Therefore $f(x',y') = f(x,y)$. Hence, since every convergent subsequence of $f(x_n,y_n)$ converge to $f(x,y)$, $\lim f(x_n,y_n) = f(x,y)$, that is, $f$ is continuous. $\blacksquare$

Is that right?

$\endgroup$
  • $\begingroup$ You can use QED in place of the black square :-) $\endgroup$ – Vim May 30 '16 at 17:24
  • $\begingroup$ Thanks ^^ Here in Brazil we use "CQD", it's good to know other versions :p $\endgroup$ – Renan R. May 30 '16 at 17:29
  • 1
    $\begingroup$ why is $f(x',y') = \lim f(x_{n_k},y')$? $\endgroup$ – user159517 May 30 '16 at 19:05
1
$\begingroup$

Several comments:

  1. It is not enough to say that $f(x_n,y_n)$ is bounded and hence it has a convergent subsequence because then you would have to write $f(x_{n_k},y_{n_k}) \rightarrow L$ and you wouldn't a priori know that $L$ has the form $L = f(x',y')$ for some $(x',y') \in \mathbb{R}^2$ like you assume in the proof. In the way your argument is written, nothing precludes the sequence to converge to a limit outside the image of $f$. However, you know something stronger - since $f(K)$ is compact and $f(x_n,y_n) \in f(K)$, there exists a subsequence $f(x_{n_k},y_{n_k})$ that converges to an element of $f(K)$ and hence the limit is of the form $f(x',y')$ for $(x',y') \in K$.
  2. Your "note that $f(x',y') = \lim f(x_{n_k},y')$" needs to be justified as this is the most important point of the proof and you rely on it heavily.
  3. In the end, I would write "the argument shows that every subsequence of $f(x_n,y_n)$ has a convergent subsequence that converges to $f(x,y)$ and so..." and not "Hence, since every convergent subsequence..." because, strictly speaking, this is enough (maybe $f(x_n,y_n)$ has subsequences that don't converge?).
$\endgroup$
  • $\begingroup$ 1. Yes yes, my fault! I forget to say that f(K) is compact. 2. It's sufficient write "f(x',y') = f(lim xnk, y') = lim f(xnk, y')"?. 3 I think that is a English problem, I didn't learn it yet :c thanks! $\endgroup$ – Renan R. May 30 '16 at 19:25
  • $\begingroup$ Regarding 2, why $f(x',y') = f(\lim_{k \to \infty} x_{n_k}, y')$? $\endgroup$ – levap May 31 '16 at 5:19
  • $\begingroup$ Well... I used that $\lim x_{n_k} = x'$, but I don't know if that is true. Any hint of how can I proceed to justify that argument? $\endgroup$ – Renan R. Jun 1 '16 at 19:25
  • $\begingroup$ Edit: That isn't true $\endgroup$ – Renan R. Jun 1 '16 at 19:26

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.