# Set Inclusion Claim in Open Mapping Theorem

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.

## 1 Answer

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 $$\|x\|=M\|y\|\|x_0\|\leq M\|y\|.$$

• Do we need $\bar{B}_X$ rather than $B_X$? – bitesizebo Apr 9 '18 at 2:50
• this was a typo in my question, I fixed it – yoshi Apr 9 '18 at 2:54
• @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. – Aweygan Apr 9 '18 at 2:56