# Linear Transformation between Topological Vector Spaces Continuous iff Continuous at Origin

## Theorem

Let $K$ be a topological field.

Let $X$ and $Y$ be topological vector spaces over $K$.

Let $T : X \to Y$ be a linear transformation.

Then $T$ is everywhere continuous if and only if it is continuous at ${\mathbf 0}_X$.

## Proof

### Necessary Condition

Suppose that $T$ is everywhere continuous.

Then $T$ is continuous at every point.

In particular, $T$ is continuous at ${\mathbf 0}_X$.

$\Box$

### Sufficient Condition

Suppose that $T$ is continuous at ${\mathbf 0}_X$.

We aim to show that $T$ is continuous at every point.

Let $x \in X$.

Let $V$ be an open neighborhood of $T x \in Y$.

We aim to show that there exists an open neighborhood $U$ of $x$ such that:

- $T \sqbrk U \subseteq V$

From Translation of Open Set in Topological Vector Space is Open, we have that:

- $V - T x$ is an open neighborhood of ${\mathbf 0}_Y$.

Since $T$ is continuous at ${\mathbf 0}_X$ and $\map T { {\mathbf 0}_X} = {\mathbf 0}_Y$, there exists an open neighborhood $U'$ of ${\mathbf 0}_X$ such that:

- $T \sqbrk {U'} \subseteq V - T x$

Then we have, by translation by $T x$:

- $T \sqbrk {U'} + T x \subseteq V$

From Image of Translation of Set under Linear Transformation is Translation of Image, we then have:

- $T \sqbrk {U' + x} \subseteq V$

Since $U'$ is an open neighborhood of ${\mathbf 0}_X$, $U = U' + x$ is an open neighborhood of $x$ with:

- $T \sqbrk U \subseteq V$

Hence the result.

$\blacksquare$

## Sources

- 1991: Walter Rudin:
*Functional Analysis*(2nd ed.) ... (previous) ... (next): $1.17$: Theorem