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I would like to demonstrate the following.

Let $X,Y$ be Banach spaces and $T:X \rightarrow Y$ be a linear operator.

Hyposthesis: Suppose you can take an open ball in Y such that $B(y, \epsilon) \underline{\subset} T(B(x, \delta))$ where $\epsilon, \delta$ are usual radii.

If we 'shift' these balls by a factor of $\alpha>0$ then we would have that the 'shifted' open y-ball would still be entirely contained in the trasnformed open Tx-ball: $B(\alpha y, \alpha\epsilon) \underline{\subset} T(B(\alpha x, \alpha\delta))$


In trying to solve this, I came up with the following:

Let us represent the open Tx-ball by $B_{Tx} = \{T(x) : x \in X \text{ and } \|x \| < \delta \}$

Since we have that $B_y \underline{\subset} B_{Tx}$, I reckon this means that the operator is bounded and possibily surjective. I am not sure how to show this though.

Since $T$ is linear then $|\alpha|\|Tx\|=\|\alpha Tx\| =\|T\alpha x\|$ and if I am right about boundedness then $\|T\alpha x\|<c_{\alpha}\|\alpha x\|$ for some scalar $c_{\alpha}$.

Now here comes the yet another issue.

The scalar multiplication on the open y-ball yields $|\alpha|\|y\| = \|\alpha y\| <|\alpha|\epsilon$. But how can infer that $B(\alpha y, \alpha\epsilon) \underline{\subset} T(B(\alpha x, \alpha\delta))$ ?

I would really appreciate your help.

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    $\begingroup$ Note that $T(B(x,\delta)) \neq B_{Tx}$, the way you've written it, unless I am misunderstanding what you are trying to say. $\endgroup$ – rubikscube09 May 19 at 18:19
  • $\begingroup$ Hey @rubikscube09, I suppose it should be like this $B_{Tx} = \{T(x) : x \in X \text{ and } \|x \| < \delta \}$ then. $\endgroup$ – upStoneLock May 19 at 18:25
  • $\begingroup$ Yes, that's correct. $\endgroup$ – rubikscube09 May 19 at 18:28
  • $\begingroup$ Great, I'll edit the text and make some adjustments. Appreciate that. $\endgroup$ – upStoneLock May 19 at 18:29
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Here's how one can go about this:

Fix $y'\in B(\alpha y,\alpha\varepsilon)$. Then it follows that $\alpha^{-1}y'\in B(y,\varepsilon)$, and since $B(y,\varepsilon)\subset T(B(x,\delta))$, there exists $x'\in B(x,\delta)$ with $Tx'=\alpha^{-1}y'$. So $y'=T(\alpha x')$, and one can check that $\alpha x'\in B(\alpha x,\alpha\delta)$, so that $y'\in T(B(\alpha x,\alpha\delta))$.

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  • $\begingroup$ Thanks a bunch. Can we say anything about it being surjective? I thought so, because we could make the image as big as we wanted by scaling it. I appreciate your help! $\endgroup$ – upStoneLock May 19 at 18:57

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