If $T: X \to Y$ is a continuous bijective linear operator where $X$ and $Y$ are Banach spaces, show that $T^{-1}$ is continuous.
Attempt:
Suppose that $T^{-1}$ is not continuous. Then, $T^{-1}$ is not bounded, so we have that $$||T^{-1}y_0||_X \geq C||y_0||_Y$$ for all $C \in \mathbb{R}$. However, since $y_0 \in Y$ and $T$ is surjective, we have that $T(x_0) = y_0$ for some $x_0 \in X.$ Then,
$$||T^{-1}(Tx_0)||_X \geq C||T(x_0)||_Y$$
$$||x_0||_X \geq C||T(x_0)||_Y$$
$$||Tx_0||_Y \leq \frac{1}{C}||x_0||_X.$$
Since this is true for all $C \in \mathbb{R}$, we have that
$$||Tx_0||_Y = 0.$$
Since we can repeat this process for every $y \in Y$, we get that
$$||Tx||_Y = 0, \ \forall x \in X.$$
Then, we must have that $Tx = 0$ for all $x$, because in general, $||x|| = 0$ iff $x=0$. So, this $T$ isn't injective, a contradiction. Thus, $T^{-1}$ must be bounded, and hence continuous.
Could someone please verify if I'm doing this properly? Linear operators are very new to me, and slightly uncomfortable. In particular, when we say a linear operator is bounded, I'm not sure if that means at a single $x_0$, or for all $x \in X$. We've also never discussed the composition of linear operators, so I'm not 100% sure that $T^{-1}T(x_0) = x_0$, although I assume it should work just like function composition. Thanks for any help in advance.
Best Answer
The proof looks good to me. To say it is ``bounded" means that there is a bound on how much a linear operator can stretch a vector.
It seems like you know the definition, but we want to just ensure that there exists some $c$ so that $\|Tx\| \leq c\|x\|$ for all $x \in X$.
Your second question is just the definition of an inverse, as in $TT^{-1}x=x$.