Let $X$, $Y$ be Banach spaces. and let $T: X \rightarrow Y$ be a bijective bounded linear map.
So I'm trying to prove that Bounded Inverse Theorem.
I already showed that for each $y \in Y$, there exists a $x\in X$ such that $Tx=y$ and $\| x\| \leq M\| y\|$, for some constant $M>0$.
Hence, we have $\| T^{-1}y\| = \|x \| \leq M\| y\|$.
Am I done here? If not, how would I finish proving this?
I'm worried that the boundedness of $\| T^{-1}y\|$ is really depending on $y$…
Thank you.
Best Answer
I think the tricky part here is that the Bounded Inverse Theorem is equivalent to other strong results such as the Open Mapping Theorem and the Closed Graph Theorem (see the first paragraph in the Wikipedia page of this theorem https://en.wikipedia.org/wiki/Bounded_inverse_theorem).
So, if you have or want to find an "easy" proof of the BIT then, in some way, you can produce an "easy" proof of the other two theorems which, if I may add, I don't think there is one (for instance, the proof I know of the OMT needs the Baire Category Theorem).
With that said, one thing I can do for you is to show you a way to prove the BIT assuming the OMT, and give you a reference where you can check the proof of the OMT. Recall that a function $f:X\to Y$ between topological spaces is called open if and only if every open subset $U$ of $X$ satisfies that $f[U]$ is an open subset of $Y$.
Open Mapping Theorem: Let $X$ and $Y$ be Banach spaces and the linear operator $T:X\to Y$ be continuous. Then its image $T[X]$ is a closed subspace of $Y$ if and only if the operator $T$ is open.
Fact: A linear operator between normed linear spaces is continuous if and only if is bounded.
Assuming the OMT and the fact, the proof of the BIT is quite easy. From the fact we deduce that, in order to check that $T^{-1}$ is bounded, it is enough to prove that $T^{-1}$ is continuous. Now, observe that $T^{-1}$ is continuous if and only if $T$ is an open map. So, if we take a look at the OMT, the only thing we have to show is that $T[X]$ is a closed subspace of $Y$. Lastly, since $T$ is a surjective function, we have the equality $T[X]=Y$; so, trivially, $T[X]$ is closed in $Y$ and the result follows.
The material discussed in this answer can be found in Royden & Fitzpatrick - Real Analysis, Fourth Edition, Chapter 13. Particularly, check Theorem 1, Theorem 8 and Corollary 9.
Hope this helps.