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Let $X$ and $Y$ be Banach spaces. Suppose $T$ is a linear operator from $X$ onto $Y$ with $\operatorname{Dom}(T)\subset X$. Show that $\exists T^{-1}\in L(Y,X)\Leftrightarrow\exists M>0:\ \left\Vert x\right\Vert \leq M\left\Vert Tx\right\Vert ,\forall x\in \operatorname{Dom}(T)$ .

This problem seems to be closely related to the following theorems. But I also noticed some differences. For example, it is possible that $\operatorname{Dom}(T)\neq X$. Also, we don't know $T\in L(X,Y)$. Thank you!

$T\in L(X,Y)$, $\operatorname{Dom}(T)=X$. Then, $\operatorname{Ker}(T)=\{0\}$ (one-to-one) and $\operatorname{Im}(T)$ is closed in $Y$ iff $\left\Vert Tx\right\Vert _{Y}\geq C\left\Vert x\right\Vert _{X}$ $\forall x\in X$ for some $C>0$.

$T\in L(X,Y)$ is bijective $\Rightarrow$$T$ is invertible. $\exists T^{-1}\in L(Y,X)$ .

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1 Answer 1

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This is fairly straight forward: $∃T^{−1}∈L(Y,X)$ obviously implies $∃M>0: ∥x∥≤M∥Tx∥,∀x∈Dom(T)$ because if $x$ is in $Dom(T)$, then $Tx=y$ for some $y\in Y$. Let $\Vert T^{-1} \Vert = M$ then it follows $$ \frac{\Vert T^{-1}(y)\Vert}{\Vert y\Vert} \leq M. $$ For the other direction we need to show that $T$ is injective. But if $T(x_1)=T(x_2)$ then, using the inequality above and the linearity of T, we get $$ {\Vert x_1 - x_2\Vert} \leq M {\Vert T(x_1) - T(x_2)\Vert} =0, $$
hence $x_1 = x_2$. So $T^{-1}$ is therefore well defined and your inequality exactly states, that it is also bounded, that is $T^{−1}∈L(Y,X)$.

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