# Functional Analysis problems

#### Oxymoron

Im trying to prove the following proposition

Let $(X,\|\cdot\|_X)$ and $(Y,\|\cdot\|_Y)$ be normed vector spaces and let $T:X \rightarrow Y$ be a surjective linear map.
Then $T$ is an isomorphism if and only if there exist $m,M > 0$ such that

$$m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X$$

For the forward inclusion, I supposed that $T$ is an isomorphism and I want to prove that $m,M > 0$ exist such that

$$m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X \quad \forall \, x \in X$$

Now since T is an isomorphism I know three things...
1) $T$ is bijective
2) $T$ bounded
3) $T^{-1}$ is bounded

So I take any $x \in X$. Then since $T$ is linear and bounded...

$$\|Tx\|_Y \leq M\|x\|_X$$

for some $M > 0$. My question is how do I show that $\|Tx\|_Y \geq m\|x\|_X$ at the same time? Do I use the inverse $T^{-1}$?

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