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