# Linear operator in banach spaces: multiplication of open balls by positive scalar

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 $$\|T\alpha x\| for some scalar $$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.

• 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. – rubikscube09 May 19 at 18:19
• Hey @rubikscube09, I suppose it should be like this $B_{Tx} = \{T(x) : x \in X \text{ and } \|x \| < \delta \}$ then. – upStoneLock May 19 at 18:25
• Yes, that's correct. – rubikscube09 May 19 at 18:28
• Great, I'll edit the text and make some adjustments. Appreciate that. – upStoneLock May 19 at 18:29

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