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Corollary 2.7 from Brezis, Functional Analysis is stated below.

Let $E$ and $F$ be two Banach spaces and let $T$ be a continuous linear operator from $E$ to $F$ that is bijective, i.e., injective and surjective. Then $T^{-1}$ is also continuous (from F into E).

Proof. Property (7) and the assumption that $T$ is injective imply that if $x \in E$ is chosen so that $||Tx|| < c$, then $||x|| < 1$. By homogeneity, we find that \begin{align*} ||x|| \leq \frac{1}{c} ||Tx|| \hspace{.5em} \forall x \in E \end{align*}

Property (7) was that if $E$ and $F$ are two Banach spaces and $T$ is a continuous linear operator from $E$ into $F$ that is surjective, then there exists some constant $c > 0$ such that $T(B_E(0,1)) \supset B_F(0,c)$.

I'm having trouble understanding the line: "By homogeneity, we find that ..."

So I tried assuming that there exists an $x \in E$ such that \begin{align*} ||x|| > \frac{1}{c} ||Tx|| \end{align*} Then we could write $c||x|| > ||Tx|| \Rightarrow c > \frac{||Tx||}{||x||}$. But since $||x|| < 1$, we then have $c > ||Tx||$. But this doesn't seem to contradict anything, and I'm not sure why homogeneity applies here.

Any suggestions would be helpful!

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    Something stronger is true: https://en.wikipedia.org/wiki/Open_mapping_theorem_(functional_analysis) – 2017-01-05

2 Answers 2

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Suppose $\|x\| > \frac{1}{c}\|Tx\|$. Then $x\ne 0$, so we can say that $\frac{\|Tx\|}{\|x\|} < c$, and in particular we can say that there exists $\epsilon>0$ such that $\frac{\|Tx\|}{\|x\|} = c-\epsilon$. Now, let $$ y = \frac{c-\epsilon/2}{\|Tx\|}x. $$ Then $$ \|y\| = (c-\epsilon/2)\frac{\|x\|}{\|Tx\|} = \frac{c-\epsilon/2}{c-\epsilon} > 1 $$ but $$ \|Ty\| = \left\|T\left(\frac{c-\epsilon/2}{\|Tx\|}x\right)\right\| = \frac{c-\epsilon/2}{\|Tx\|}\|Tx\| = c-\epsilon/2 < c $$ so $y$ satisfies $\|Ty\| < c $ but $\|y\| > 1$, contradicting the assumption that $\|Ty\|

We can also argue this directly, as follows: suppose $x\ne 0$, so that $Tx\ne 0$ by injectivity. We know that for any $\alpha>0$, we have $$\|x\| = \frac{\|\alpha x\|}{\alpha} = \frac{\|\alpha x\|\|Tx\|}{\alpha\|Tx\|} = \frac{\|\alpha x\|}{\|T(\alpha x)\|}\|Tx\|.$$ The ratio $k = \frac{\|\alpha x\|}{\|T(\alpha x)\|}$ is a constant independent of $\alpha$. Now, if we choose $\alpha$ small enough so that $\|T(\alpha x)\|0$, we have that $$\|T(\alpha x)\|

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So, we know from the open mapping theorem, $$(\exists c > 0)\quad ||Tx|| < c \implies ||x||<1.$$ It follows that if $||x|| \geq 1,$ then $||Tx|| \geq c$.

For $x=0$ the statement is true (because then $||Tx||=||x||=0$), so let $x \neq 0$. $$||\frac{x}{||x||}|| \geq 1,$$ so $$||\frac{Tx}{||x||}|| \geq c,$$ from which the result follows.