Prove that $T$ is continuous if and only if $N_T=\{0\}$

Question

Let $X$ and $Y$ be a Banach spaces and let $T : X \to Y $ be a linear operator. Consider the set $$N_T := \bigcap \{\overline{T(V)} : V \text{ is a neighborhood of } 0\},$$ where $\overline{A}$ denotes the closure of a subset $A \subset Y$. Notice $N_T$ is a closed linear subspace of $Y$.

First I'd like to point out that this set seems to remind me of the kernel of a linear operator, but clearly this isn't the case, since $N_T\subset Y$.

I'd like to show that $T$ is continuous if and only if $N_T = \{ 0 \}$.

First we can see that $0\in N_T$, since for every neighbourhood V of 0, we have that $T(0)\in\overline{T(V)}$, and by linearity we have that $T(0)=0$, thus, $0\in N_T$. I also tried using the fact that $T$ is continuous iff $T$ is continuous at 0, but that didn't take me very far either. Apart from showing that I can't seem to go any further. I'm having a lot of problems even understanding what this set is to begin with. For example, I'm not even sure that for every $n\in N_T$ there exists an $x\in X$, such that $Tx=n$, since we're taking the intersection of $\overline{T(V)}$ and not just $T(V)$ itself.

I would really appreciate any help. Thank you.

Answer 1

I'll give hints to both directions.

Suppose $T$ is continuous, and let $y\in N_T$. If $B(0,\epsilon)$ denotes the open ball centered at $0$ with radius $\epsilon$, then for each $n\in\mathbb{N}$ we can choose $x_n\in B(0,\frac{1}{n})$ such that $||T(x_n)-y||<\frac{1}{n}$. Can you see how to conclude $y=0$?

For the converse, assume $N_T=\{0\}$. Since the spaces are Banach, we may use the closed graph theorem. So assume $x_n\to 0$ and $T(x_n)\to y$. Our aim is to show that $y=T(0)=0$. To do so, show that $y\in N_T$, which is pretty straightforward.

Answer 2

We want to show that $T$ is continuous if and only if $N_T=\{0\}$.

$(\impliedby):$ Suppose $N_T=\{0\}$, and let $\{x_j\}_{j\in\mathbb{N}}$ be a sequence in $X$ converging to $0$ such that $\{Tx_j\}_{j\in\mathbb{N}}$ converges to $y\in Y$. If we show that $y=0$, then it follows by the closed graph theorem that $T$ is continuous, so we do this. Observe that for any neighborhood $V$ of $0$ (in $X$) we can find some $N\in\mathbb{N}$ such that $x_j\in V$ for all $j\geq N$. In particular $Tx_j\in T(V)$ for all $j\geq N$. This implies that $y\in\overline{T(V)}$, and as $V$ was arbitrary, $y\in N_T$. Consequently $y=0$, and the result follows.

$(\implies):$ Suppose now that $T$ is continuous, and let $y\in N_T$. In particular $y\in\overline{T(B_{2^{-j}}(0))}$ for all $j\in\mathbb{N}$, and so for each $j\in\mathbb{N}$ we can find some $x_j\in B_{2^{-j}}(0)$ such that $\lVert Tx_j-y\rVert<\frac{1}{2^j}$. But now observe that $x_j\to0$ as $j\to\infty$, so by continuity $Tx_j\to0$ as $j\to\infty$. It follows that

$$0\leq \lVert y\rVert=\lim_{j\to\infty}\lVert Tx_j-y\rVert\leq\lim_{j\to\infty}\frac{1}{2^j}=0,$$

so $y=0$. It follows that $N_T\subseteq\{0\}$. As we clearly always have that $0\in N_T$, the result follows.