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 $T^{-1}$?

 

[anathema]

Thank you for your interesting proposition and for reaching out for assistance in proving it. I would like to offer my insights on how you can approach this problem.

Firstly, your understanding of the properties of an isomorphism is correct. In order to prove the proposition, we need to show that if T is an isomorphism, then there exist m,M > 0 such that the given inequality holds for all x \in X. And if the inequality holds, then T must be an isomorphism.

Now, to show that \|Tx\|_Y \geq m\|x\|_X, we can use the inverse T^{-1} as you suggested. Since T is bijective, we know that T^{-1} exists and is also linear and bounded. This means that for any y \in Y, we can write y = T^{-1}(Tx). Using the boundedness of T^{-1}, we can write

\|y\|_Y = \|T^{-1}(Tx)\|_Y \leq M \|Tx\|_Y

where M is the bound for T^{-1}. Now, using the given inequality, we have

\|y\|_Y \leq M \|Tx\|_Y \leq M(M^{-1})\|Tx\|_Y = \|Tx\|_Y

So we have shown that for any y \in Y, there exists x \in X such that \|y\|_Y \leq \|Tx\|_Y. This implies that \|Tx\|_Y \geq \|y\|_Y, which in turn implies that \|Tx\|_Y \geq m\|x\|_X, where m = 1. Therefore, we have shown that m\|x\|_X \leq \|Tx\|_Y \leq M\|x\|_X for all x \in X.

I hope this helps in proving your proposition. Remember, as scientists, it is important to provide a rigorous and logical proof for our propositions. Good luck!

 

[quicknote]

Yes, you can use the inverse T^{-1} to show the other direction. Since T^{-1} is also bounded, there exists a constant N > 0 such that \|T^{-1}y\|_X \leq N\|y\|_Y for all y \in Y. Therefore, for any x \in X, we have \|x\|_X = \|T^{-1}Tx\|_X \leq N\|Tx\|_Y. This implies that \|Tx\|_Y \geq \frac{1}{N}\|x\|_X. Thus, we can choose m = \frac{1}{N} and have m\|x\|_X \leq \|Tx\|_Y for all x \in X. Therefore, we have shown that both inequalities hold and m,M > 0 exist, proving the proposition.