Functional Analysis – Is the Kernel of a Continuous Linear Operator a Closed Subspace?

Question

If $V$ and $W$ are topological vector spaces (and $W$ is finite-dimensional) then a linear operator $L\colon V\to W$ is continuous if and only if the kernel of $L$ is a closed subspace of $V$.

Why is this so?

Answer 1

You need the hypothesis that $V$ and $W$ be Hausdorff, although this is usually given for granted.

One direction is as in Siminore's comment.

For the other, since $\text{ker}L$ is closed, the quotient space $V/\text{ker}L$ is Hausdorff. Algebraically $V/\text{ker}L\cong\text{im}L,$ and $\text{im}L$ is a subspace of $W.$ The algebraic isomorphism $V/\text{ker}L\cong\text{im}L$ derived from $L$ is a topological isomorphism because finite-dimensional spaces admit only one Hausdorff topological vector space topology (any linear isomorphism between finite-dimensional Hausdorff topological vector spaces is a topological isomorphism). This means that $L$ is continuous, as a composition of the quotient map $V\to V/\text{ker}L$ with the isomorphism between $V/\text{ker}L$ and $\text{im}L$ with the inclusion $\text{im}L\to W.$