Define the bounded linear operator $T : L^2[0,1] \to L^2[0,1]$ by $$Tf(x) = xf(x), \quad \forall f \in L^2[0,1].$$
Let $B$ the open unit ball in $L^2[0,1]$. I need to determine whether $0$ is in the interior of $T(B)$ with respect to metric space $Y = L^2[0,1]$ or not.
Besides, there is a question that I think it may relate to the problem. I also need to prove for all $\varepsilon > 0$, there is a function $f \in L^2[0,1]$ such that $\| f \|_2 \ge 1$ but $\| Tf \|_2 < \varepsilon$.
Any help is really appreciated.
Now, I have figured out it.
First note that $T$ is injective. Let $V$ be an open set in $X$ containing $0$. Then, there is some $\varepsilon > 0$ such that $B(0,\varepsilon) \subset V$. By the fact, there is a function $f \in L^2[0,1]$ such that $f \notin B(0,1)$ but $T(f) \in B(0,\varepsilon)$. But since $T$ is injective and $f \notin B(0,1)$, $T(f) \notin T(B)$. Therefore we have $$T(f) \in V \cap Y \setminus T(B),$$ which means the set is nonempty. Hence, $0$ is not in the interior of $T(B)$.