I was following the proof of the Open Mapping Theorem in Lang's Real and Functional Analysis and something odd happened. I was able to simplify a lot his proof. Not only that, but I was able to improve the theorem itself. My proof is valid for any normed space, not only Banach spaces. Of course I know this can't be the case, but still, I wasn't able to find where my proof got wrong. If you spot the mistake, please tell me.

Open Mapping Theorem: Let $X,Y$ be Banach spaces and let $f:X \to Y$ be a linear continuous map. If $f$ is surjective, then it is also open.

Proof: For any $s>0$, we will denote $B_s = \{ x\in X:\ \|x\| < s \}$ and $C_s = \{ y \in Y:\ \|y\| < s \}$.

First, note that since $X$ is surjective, we have $$ f(X) = f(\cup_{n=1}^\infty B_n) = Y.$$ Also, note that $s < t \implies B_s \subset B_t$. In particular, this means that $\cup_{n=1}^N B_n \subset B_t$, for any $t > N$. Then for any $r > 0$, there is a $k \in \mathbb{N}$ such that $C_r \subset f(B_{kr})$.

Now let $U \subset X$ be an open set and let $x \in U$ be an arbitrary point. By definition, there is a $t > 0$ such that $x + B_t \subset U$. Then $$f(x+B_t) = f(x) + f(B_t) \subset f(U).$$

Given any $r>0$, we know there is a $k \in \mathbb{K}$ such that $C_r \subset f(B_{kr})$. The it follows that $$f(B_t) = f\left(\frac{t}{kr} B_{kr}\right) = \frac{t}{kr} f(B_{kr}) \supset \frac{t}{kr} C_r = C_\frac{t}{k}.$$

Therefore, $$f(x) + C_\frac{t}{r} \subset f(x) + f(B_t) \subset f(U). $$

This proves $f(x)$ is in the interior of $f(U)$. Since $x$ was arbitrary, we conclude that every point of $f(U)$ is in its interior, so $f(U)$ is open. $\hspace{1cm}\square$

Thank you!

  • $\begingroup$ Can i suggest a title change to "Incorrect proof of the Open Mapping Theorem." It would seem to be better for archival purposes on the site, and, after all, mistakes in our own proofs always tend to be difficult for us to spot (otherwise we probably wouldn't have made them in the first place!). $\endgroup$ – David Mar 4 '17 at 14:37
  • $\begingroup$ No problem. :)) $\endgroup$ – Integral Mar 4 '17 at 14:47

$f(\bigcup_n B_n)=Y$ does not implies there exists $l$ such that $C_r\subset f(B_l)$, the image of a ball is not always a ball.

  • $\begingroup$ I didn't say for any. What I said is that given $r>0$, there is a $k \in \mathbb{N}$ such that $C_r \subset f(B_{kr})$. $\endgroup$ – Integral Mar 4 '17 at 13:51
  • $\begingroup$ Take $kr=l$, you have $C_r\subset f(B_l)$. $\endgroup$ – Tsemo Aristide Mar 4 '17 at 13:52
  • $\begingroup$ Still, what I said translates to there is $l > 0$ such that $C_r \subset f(B_l)$. Not any $l>0$. $\endgroup$ – Integral Mar 4 '17 at 13:53
  • $\begingroup$ Why such a $l$ exists? $\endgroup$ – Tsemo Aristide Mar 4 '17 at 13:54
  • 1
    $\begingroup$ Yes, this is what I pointed out in my answer, I said that the image of a ball is not always a ball, because is that was true, your argument would have been true too. $\endgroup$ – Tsemo Aristide Mar 4 '17 at 14:01

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.