Let $X$ and $Y$ be topological vector spaces and $T\colon X\to Y$ linear. Suppose that $T$ sends bounded sets to bounded sets and that $X$ is first countable.

The claim is that $T$ is continuous.

Here's what I've tried so far. Since $X$ is first countable, we need only show that $T$ is sequentially continuous and since $T$ is linear it suffices to show $T$ is continuous at $0$. Let $(x_n)$ be a sequence in $X$ with $x_n\to 0$. We want $Tx_n\to 0$ as well. $(x_n)$ is convergent, so $\{x_n:n\in\mathbf{N}\}$ is bounded in $X$. Hence $\{Tx_n:n\in\mathbf{N}\}$ is bounded in $Y$. Let $W$ be a neighborhood of $0$ in $Y$. Then there is $s>0$ with $\{Tx_n:n\in\mathbf{N}\}\subset sW$. So, $\{x_n:n\in\mathbf{N}\}\subset sT^{-1}(W)$.

But I'm not sure how to/if I can continue successfully from here. Any thoughts?

  • $\begingroup$ As argued in my answer below, the missing component to your proof is that there are positive scalars $\lambda_{n}\rightarrow\infty$ such that $\lambda_{n}x_{n}\rightarrow 0$ in $X$. $\endgroup$ – Matt Rosenzweig Apr 3 '15 at 14:28

Lemma 1 If $X$ is first countable and $x_{n}\rightarrow 0$, then there exist positive scalars $\lambda_{n}\rightarrow\infty$ such that $\lambda_{n}x_{n}\rightarrow 0$.

Proof: Let $\left\{U_{k}\right\}$ be a balanced local base for $0\in X$ such that $U_{k+1}\subset U_{k}$. Since $x_{n}\rightarrow 0$, there exists an increasing sequence $(N_{k})$ of natural numbers such that $n\geq N_{k}$ implies $x_{n}\in k^{-2}U_{k}$. Define the scalars $\lambda_{n}$ by

$$\lambda_{n}=k \text{ if } N_{k}\leq n < N_{k+1}$$

Then $\lambda_{n}x_{n}\in k^{-1}U_{k}\subset U_{k}$ if $N_{k}\leq n < N_{k+1}$, whence $\lambda_{n}x_{n}\rightarrow 0$. $\Box$

Lemma 2 If $T$ is continuous at one point in $X$, then $T$ is continuous on $X$.

Proof: Since translation is a homeomorphism, we may assume without loss of generality that $T$ is continuous at $0\in X$. Fix $x_{0}\in X$, and let $U$ be an open nbhd of $Tx_{0}$. Then $V:=U-Tx_{0}$ is an open nbhd of $0\in Y$, whence by our hypothesis there exists an open nbhd $W$ of $0\in X$ such that

$$(x\in W\Rightarrow Tx\in V)$$

But this implies that $W':=W+x_{0}$, which is an open nbhd of $x_{0}$, satisfies

$$(x'\in W'\Rightarrow Tx'\in U)$$


In what follows below, $\left\{U_{n}\right\}$ denotes a countable local base of balanced open sets for the origin in $X$.

Theorem If $T:X\rightarrow Y$ is bounded and $X$ is first countable, then $T$ is continuous.

Proof: By the lemma, it suffices to show that $T$ is continuous at $0\in X$. Let $x_{n}\rightarrow 0$ in $X$ and let $V$ be a balanced open nbhd containing $0\in Y$ such that $Tx_{n}\notin V$ for every $n$.

By Lemma 1, there exist positive scalars $\lambda_{n}\rightarrow\infty$ such that $\lambda_{n}x_{n}\rightarrow 0$. Whence, the set $E:=\left\{T(\lambda_{n}x_{n}) : n\in\mathbb{N}\right\}$ is bounded. So there exists a scalar $s>0$ such that $\lambda_{n}Tx_{n}=T(\lambda_{n}x_{n})\in sV$ for all $n$, which implies that

$$Tx_{n}\in \dfrac{s}{\lambda_{n}}V \ \forall n$$

But for all $n$ sufficiently large $s\lambda_{n}^{-1}\leq 1$, whence $Tx_{n}\in V$, which is a contradiction. $\Box$


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