Why is $f$ continuous if its kernel is not dense in $A$?

Question

Let $A$ be a commutative complex Banach algebra with unit element $e$.

Suppose $X \subset A$ has codimension $1$ and consists out of non-invertible elements.

Clearly $X$ is the kernel for some linear functional $f$. Since $X$ contains no invertible elements, it cannot be dense in $A$.

Now my question is the following: How does the continuity of $f$ follow from this?

In the paper that I'm reading it says that $f$ is continuous directly after saying $X$ is not dense in $A$.

I know that a linear functional $f$ is continuous if and only if its kernel (which in this case is $X$) is closed. Can I somehow prove, using my given information above, that $X$ is closed? This will then lead to the continuity of $f$, but I cannot see how the fact that $X$ is not dense in $A$ leads to us knowing that $X$ is closed.

Can anyone please help provide me with some clarity?

Answer 1

The closure of a subspace is a subspace; since $X$ has codimension $1$, the only subspace properly containing it is $A$. Since it is not dense, its closure is not $A$. So the closure of $X$ is $X$.

A functional with closed kernel is continuous.