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Let $X,Y$ be Banach spaces and let $A:X\to Y$ by a bijective bounded linear operator. Let $\delta > 0$ be a constant such that $B_\delta(0;Y) \subset A(B_1(0))$. Then we have that $$ \inf_{\substack{x\in X\\ Ax = y}} ||x||_X \le \delta^{-1}||y||_Y, \quad \text{for all} \ y \in Y. $$ I want to show that the inverse operator $A^{-1}:Y\to X$ is bounded by showing that $||A^{-1}|| \le \delta^{-1}$.

Here is what I have done: $$ \begin{align} ||A^{-1}|| = \sup_{||y||=1}||A^{-1}y||_X \\ = \sup_{{\substack{x\in X\\ Ax = y\\||y||=1}}}||x||_X \\ \end{align} $$ But I have a $\sup$ in this expression instead of an $\inf$ so I can't use the inequality above to finish the proof?

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We only use $B_\delta(0;Y) \subset A(B_1(0))$:

to this end take $y \in Y$ with $y \ne 0$. Then $z:=\frac{\delta}{||y||_Y}y \in B_\delta(0;Y) \subset A(B_1(0))$. Hence there is $x \in X$ with $||x||_X \le 1$ and $z=Ax$. It follows:

$A^{-1}y=\frac{||y||_Y}{\delta}x$, hence

$$||A^{-1}y||_X \le \frac{||y||_Y}{\delta}.$$

This gives

$$||A^{-1}|| \le \delta^{-1}.$$

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  • $\begingroup$ That's great thanks! Is it also possible to obtain it with the method I was using or had I run into a 100% dead end? $\endgroup$ – ManUtdBloke Jul 29 '17 at 8:32

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