1
$\begingroup$

I'm reading Royden and trying to understand some aspects of the open mapping theorem's proof.

He begins the section with the following theorem:

Let $X$ and $Y$ be Banach spaces and the linear operator $T:X \rightarrow Y$ be continuous. Then $T(X)$ is a closed subspace of $Y$ if and only if there is a constant $M > 0$ for which given $y \in T(X)$, there is an $x \in X$ such that $T(x) = y$ and $$\|x\|\leq M \|y\| \text{ (*)}$$.

In the proof of the open mapping theorem he then states:

($B_x$ and $B_y$ are unit balls at the origin of $X$ and $Y$, resp) We infer from the homogeneity of $T$ and of the norms that (*) is equivalent to the inclusion $$\overline{B_Y} \cap T(X) \subseteq M\text{ }T(\overline{B_X})$$

This confuses me because the inequality in (*) is between norms on two different spaces and the inclusion on the bottom is all in $Y$. So it's not clear to me how to go from one to the other.

$\endgroup$
0
1
$\begingroup$

Suppose first that there is an $M>0$ such that $(*)$ holds. If $y\in T(X)$ and $\|y\|\leq1$, then there is some $x\in X$ with $Tx=y$ and $\|x\|\leq M\|y\|\leq M$. Thus we have $$\overline B_Y\cap T(X)\subset T(M\overline B_X)=M\ T(\overline B_X).$$ Suppose now that $\overline{B_Y} \cap T(X) \subseteq M\ T(\overline B_X)$. Fix $y\in T(X)$. If $y=0$, we can choose $x=0$ so that $(*)$ holds, so suppose $y\neq 0$. Then $\frac{1}{\|y\|}y\in\overline{B_Y} \cap T(X)$, so there is some $x_0\in \overline B_X$ such that $\frac{1}{\|y\|}y=M\ Tx_0$. Put $x=M\|y\|x_0$. Then $y=Tx$, and $$\|x\|=M\|y\|\|x_0\|\leq M\|y\|.$$

$\endgroup$
3
  • 1
    $\begingroup$ Do we need $\bar{B}_X$ rather than $B_X$? $\endgroup$ – bitesizebo Apr 9 '18 at 2:50
  • 1
    $\begingroup$ this was a typo in my question, I fixed it $\endgroup$ – yoshi Apr 9 '18 at 2:54
  • 1
    $\begingroup$ @yoshi I susupected it was either that or one of the books numerous errors. I'm going to bed now, if you have any questions just leave a comment and I'll get back to you tomorrow. $\endgroup$ – Aweygan Apr 9 '18 at 2:56

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.