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.

  • 1
    $\begingroup$ 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

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 $y'\in T(B(\alpha x,\alpha\delta))$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.