# Linear homeomorphism between normed linear spaces

I want to prove the following exercise:

Let $L:V\to W$ be a linear map between two normed linear spaces such that $L(V)=W$. Prove that $L$ is a linear homeomorphism if and only if there exist $m,M>0$ such that $$m\Vert x \Vert \leq \Vert Lx \Vert \leq M \Vert x \Vert$$ for all $x\in V$.

I have the definition of homeomorphism ($f: X\to Y$ continuous bijection is a homeomorphism if $f^{-1}$ is continuous) and the following equivalent propositions:

1. $f$ is a homeomorphism
2. $f$ is open and continuous
3. $f$ is closed and continuous
4. $f(\overline A) = \overline{f(A)}$ for all $A\subset X$ ($\overline A$ is the closure of $A$)

I think this is all the information about homeomorphisms I have on my notes, but I don't see the relation between this and the fact that $L$ is bounded (which is what $m\Vert x \Vert \leq \Vert Lx \Vert \leq M \Vert x \Vert$ means).

Suppose $L$ is a homeomorphism. Then $L$ is bounded, and hence $\| L x \| \leq \|L \| \|x \|$. Similarly, $\|x\| = \| L^{-1} L x \| \leq \| L^{-1} \| \|L x \|$. Setting $m := \|L^{-1} \|^{-1}$ and $M := \|L\|$ yields $$m \| x \| \leq \| L x \| \leq M \| x \|.$$

Conversely, suppose there exists $m, M > 0$ such that, for all $x \in V$, $$m \| x \| \leq \| L x \| \leq M \| x \|.$$ Then, in particular, it holds that $\| L x \| \leq M \| x \|$ for all $x \in V$, that is, the linear map $L$ is bounded. Furthermore, the inequality $m \| x \| \leq \| L x \|$ shows that $L x \neq 0$ whenever $x \neq 0$, that is, the map $L$ is injective. As you have the surjectivity of $L$ by assumption, it follows that $L$ is bijective. To show that $L^{-1}$ is bounded, note that $\|L^{-1} L x \| = \|x \| \leq m^{-1} \|Lx\|$ for all $x \in X$.

You don't need to use any of your equivalent conditions, since this is easier working from the original definition of a homeomorphism. Here is the sketch:

• Prove that $L$ is continuous. If you know the following useful fact, then you can use it:

If $X,Y$ are normed spaces and $T\colon X\to Y$ is a linear map then $T$ is continuous if and only if it is bounded; i.e., there is some $C$ such that $\|Tx\|\le C\|x\|$ for all $x\in X$.

• It is given that $L\colon V\to W$ is surjective. Use the given inequality to prove that $L$ is injective as well.

• Then $L$ has an inverse $L^{-1}\colon W\to V$. Prove that $L^{-1}$ is a linear map.

• Prove that $L^{-1}$ is continuous. You can use the same argument that you used to show that $L$ is continuous.

• Now we prove the other direction. Suppose that $L\colon V\to W$ is a homeomorphism. Then $L$ and its inverse $L^{-1}$ are both bounded. Use this to derive the inequality.