Let T be a linear operator from normed spaces X to Y if $m\|x\|\leq \|T(x)\|$ for $m>0$ then $T^{-1}$ exists and continuous.

Question

The proof I'm looking at is from page 6 of https://courses.smp.uq.edu.au/MATH3402/Lectures/normls.pdf and begins half way down the page.

The proof claims invertibility without showing surjectivity. Is there anyway to prove T is surjective from the hypothesis or is this some assumption? Aside from this point I understand the proof.

Answer 1

I think we should assume that $T$ from $X$ to $Y$ here means that $T$ has $Y$ as its image space, i.e. is already given to be onto. Using onto in that sentence already would have clarified that.

It says "The inverse $T^{-1}$ exists and is continuous iff there is a constant $m>0$ such that $m\|x\| \le \|T(x)\|$ for every $x \in X$.

We cannot conclude only from the existence of this constant that $T$ is invertible, e.g. use $T(x)= (x,0)$ from $\Bbb R$ to $\Bbb R^2$, where $m=1$ works. So the ontoness is implied by the $T^{-1}$ existing, and the proof of injectivity then becomes superfluous.

As it stands the statement is false though.

The fact could have better been formulated as

Let $T:X \to Y$ be a continuous linear map. If $T^{-1}$ exists, it is also continuous, and hence there are constants $m,M>0$ such that $m\|x\|\le \|T(x)\| \le M\|x\|$ for all $x \in X$.

Answer 2

As stated in the linked notes, the assertion is wrong. You need that $T$ is surjective which does not follow from the assumption.

I guess the author made a typo. He should have written

Let $T$ be a linear operator from $X$ onto $Y$.