Why $\overline{D(Y) \cap T(D(X))} = B(Y)$ holds?

Question

For any normed space $X,$ let $D(X)$ and $B(X)$ be the open and closed unit ball of $X$ centered at origin respectively.

The following lemma is obtained from Bollobas's Linear Analysis, page $79.$

Lemma: Let $X$ and $Y$ be normed spaces, $X$ is complete and $T:X\to Y$ is a bounded linear operator. If $$D(Y)\subseteq \overline{T(D(X))},$$ then $$D(Y)\subseteq T(D(X)).$$

The proof starts with the following equation $$\overline{D(Y) \cap T(D(X))} = B(Y).$$ Clearly $D(Y) \cap T(D(X)) \subseteq D(Y)\subseteq B(Y).$ Therefore, $$\overline{D(Y)\cap T(D(X))}\subseteq B(Y).$$ However, I do not understand why the reverse inclusion holds.

Any hint would be appreciated.

Answer 1

For $\|y\|=1$, consider $y_{n}=\dfrac{y}{\|y\|+(1/n)}=\dfrac{y}{1+1/n}\in D(Y)$, then $\|y_{n}-T(x_{n})\|<\dfrac{1}{2(n+1)}$ for some $x_{n}\in D(X)$. Now \begin{align*} \|T(x_{n})\|\leq\|T(x_{n})-y_{n}\|+\|y_{n}\|<\dfrac{1}{2(n+1)}+\dfrac{n}{n+1}<1, \end{align*} and \begin{align*} \|T(x_{n})-y\|\leq\|T(x_{n})-y_{n}\|+\|y_{n}-y\|<\dfrac{1}{2(n+1)}+\dfrac{1}{n+1}\rightarrow 0. \end{align*} But $T(x_{n})\in D(Y)\cap T(D(X))$.