In Haim Brezis's book "Functional Analysis," the page $35$ writes the corollary $2.7$:
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
(= one-to-one) and surjective. Then $T^{-1}$ is also continuous (from
$F$ into $E$).
The proof goes like this and I will parapharse it a little bit to make it simpler.
Proof. By Open Mapping Theorem and injectivity of $T$, it implies whenever $x \in E$ with $\|Tx\| <c,$ we will have $\|x\| < 1.$ By homogeneity, we find that
$$\|x\| \leq \frac{1}{c} \|Tx\| \quad \forall x \in E$$
and therefore $T^{-1}$ is continuous.
Here is my doubt: why we can use homogeneity here? What does it really mean? How we use homogeneity in other cases? (group theory, ring theory, or category theory)
Best Answer
He simply means $\forall \lambda\; \lambda Tx=T(\lambda x)$. Since if you know $$\Vert Tx \Vert \leq c \implies \Vert x \Vert \leq 1$$ (I use weak inequalities, since then it's more accurate than what is claimed in the OP) Then by Homogeneity, for each $x \in E$ we get ($\lambda = \frac{c}{\Vert Tx \Vert})$: $$ \Vert T\big(c\frac{x}{\Vert Tx \Vert}\big) \Vert \leq c \implies \Vert c \frac{x}{\Vert Tx \Vert} \Vert \leq 1 \implies \Vert x \Vert\leq \frac{1}{c} \Vert Tx \Vert$$