Talk:Kernel of Bounded Linear Transformation is Closed Linear Subspace

Add the converse

I suggest changing the statement to "if and only if", that is

$\ker f$ is closed if and only if $f$ is continuous.

For the converse, assume that $f\neq0$ and that $\ker f$ is a closed set.

Pick some $e$ in $X$ with $\map f e = 1$.

Suppose by way of contradiction $\norm f = \infty$.

Then there exists a sequence $\sequence {x_n}$ in $X$ with $\norm {x_n} = 1$ and $f(x_n)\ge n$ for all $n$.

Note that the sequence $\sequence {y_n}$ defined by

$y_n = e - \dfrac{x_n}{\map f {x_n}}$

satisfies $y_n \in \ker f$ for all $n$ and $y_n \to e$.

Since the set $\ker f$ is closed it follows that $e$ must belong to it

and consequently $\map f e = 0$ which is a contradiction.

Thus $f$ is a continuous linear functional.

--Hbghlyj (talk) 18:19, 17 March 2024 (UTC)