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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.

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  • $\begingroup$ This isn't right. Note your first display says $T^{-1}y_0$ has infinite norm... But, try using the Open Mapping Theorem to show $(T^{-1})^{-1} U$ is open for any $U$ open in $X$. $\endgroup$ – David Mitra Sep 24 '17 at 20:23
  • $\begingroup$ Oh. It's literally just, when $U \subset X$ is open, then since $T$ is a continuous surjective linear operator, by the Open Mapping Theorem, $T(U)$ is open, which is precisely what $T^{-1}$ being continuous means? So... $T^{-1}y_0$ can't have infinite norm? Isn't this just the negation of being bounded? $\endgroup$ – user389056 Sep 24 '17 at 23:02
  • $\begingroup$ If $y_0$ is fixed, no; $T^{-1}y_0$ is some element of $X$, it has finite norm. But $\sup_{\Vert y \Vert=1} \Vert T^{-1} y\Vert$ can be infinite. Different $C$ correspond to different $y$. $\endgroup$ – David Mitra Sep 25 '17 at 6:22
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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 $TT^{-1}x=x$.

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  • $\begingroup$ So, the same $c$ for all $x$? Okay, thanks. $\endgroup$ – user389056 Sep 24 '17 at 19:11
  • $\begingroup$ Follow up question, if you have a chance: If we say that a family $T_j$ is bounded, then $||T_j(x)|| < C||x||$ for all j and x, or is every j allowed to use a different C? $\endgroup$ – user389056 Sep 24 '17 at 19:21
  • $\begingroup$ Look up the uniform boundedness principle. I think it depends on the context, but more likely than not, the former $\endgroup$ – Andres Mejia Sep 24 '17 at 19:44
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By open mapping theorem, the preimage of open set in $X$ under $T^{-1}$ is still open in $Y$, which shows that $T^{-1}$ is continuous.

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