Prove that $\forall \delta>0,\delta B_Y\subseteq T(B_X)\Leftrightarrow \gamma \delta B_Y \subseteq\gamma T(B_X) \quad \forall \gamma >0 $

Question

Let $X,Y$ normed vectorial space and $T\in B(X,Y).$ Prove that $$\forall \delta>0,\delta B_Y\subseteq T(B_X)\Leftrightarrow \gamma \delta B_Y \subseteq\gamma T(B_X) \quad \forall \gamma >0 $$

My attempt:

If $\quad \forall \delta >0 \quad\delta B_Y \subseteq T(B_X)$ then $\gamma \delta B_Y \subseteq T(\gamma B_X)= \gamma T(B_X) \quad \forall \gamma >0$.

but for $\Leftarrow$?

Is it true also if I consider the clousure i.e.$\delta B_Y\subseteq \overline {T(B_X)}\Leftrightarrow \gamma \delta B_Y \subseteq \overline{\gamma T(B_X)}$?

My attemp:

$\Rightarrow$: $T(B_X)\subseteq \overline{T(B_X)}$ so $\delta B_Y \subseteq \overline{TB_X}$.Then $\gamma \delta B_Y \subseteq \gamma \overline{TB_X} \subseteq \overline{\gamma TB_X}=\overline{T(\gamma B_X)}$?

Answer 1

In particular, take $\gamma=1...$

Edit: In your second question, the closure makes no difference. If $A \subseteq B,$ then, of course, $\gamma A \subseteq \gamma B.$ And the converse is again by taking $\gamma=1.$