Linear Operator and Open Balls in a range

Question

Let $X$ and $Y$ be two Banach spaces under the same field $K = \mathbb{R}$ or $K = \mathbb{C}$, and $T: X \rightarrow Y$ is a linear operator. Show that

(a) if $B(y, \epsilon) \subset T(B(x,\delta))$, then $B(\alpha y, \alpha \epsilon) \subset T(B(\alpha x,\alpha \delta))$, where $\alpha >0$.

(b) if $B(y, \epsilon) \subset \overline{T(B(x,\delta))}$, then $B(\alpha y, \alpha \epsilon) \subset \overline{T(B(\alpha x,\alpha \delta))}$, where $\alpha >0$.

I have doubts in both cases. I read the lemma in Kreyszig's book to understand the Open Mapping Theorem, but I couldn't solve this problem and I don't understand why $X$ and $Y$ need to be Banach spaces, that is, the result continues valid for $X$ and $Y$ normed spaces?

Answer 1

The result is true if $X,Y$ are only assumed to be normed spaces (in fact, this assumption could be weakened even further). The key thing to notice is that $B(\alpha y, \alpha \varepsilon) = \alpha B(y,\varepsilon)$ and so $B(y, \varepsilon) \subseteq T(B(x, \delta))$ implies that $$B(\alpha y, \alpha \varepsilon) = \alpha B(y,\varepsilon) \subseteq \alpha T(B(x, \delta)) = T(\alpha B(x, \delta)) = T(B(\alpha x, \alpha \delta)).$$ Similarly, if $B(y, \varepsilon) \subseteq \overline{T(B(x, \delta))}$ then $$B(\alpha y, \alpha \varepsilon) = \alpha B(y,\varepsilon) \subseteq \alpha [\overline{T(B(x, \delta))}] = [\overline{\alpha T(B(x, \delta))}] = \overline{T(\alpha B(x, \delta))} = T(B(\alpha x, \alpha \delta))$$ where we conclude that $\alpha \overline{T(B(x, \delta))} = \overline{\alpha T(B(x, \delta))}$ since for fixed $\alpha \neq 0$, $ z \mapsto \alpha z$ is a homeomorphism and so we can take it inside the closure.