Prove that $\forall \delta>0,\delta B_Y\subseteq T(B_X)\Leftrightarrow B_Y(Tx_0,\delta) \subseteq T(B_X(x_0,1)) \forall x_0\in X$

Question

Let $X,Y$ normed vectorial space and $T \in B(X,Y)$. Prove that $$\forall\ \delta>0,\delta B_Y\subseteq T(B_X)\iff B_Y(Tx_0,\delta) \subseteq T(B_X(x_0,1))\; \forall\ x_0\in X$$

Here $\delta B_Y=\{y\in Y:||y||<\delta\}$.

Attempt:

$\Rightarrow$

\begin{align*}\delta B_Y &\subseteq T(B_X)\\ \implies B_Y(Tx_0,\delta)&=Tx_0+\delta B_Y\\&\subseteq Tx_0+T(B_X)\\&=T(x_0+B_X)\\&=T(B_X(x_0,1))\end{align*}

But for $\Leftarrow$?

Answer 1

The converse implication is easier.

Hint: Since it is true for all $x_0 \in X,$ it is, in particular, true for $x_0=0.$