Closed linear operators have closed kernels

Question

Let $X,Y$ be Banach spaces and $K\colon X\to Y$ be a closed linear operator. I would like to show that its kernel is closed subspace of $X$. Any ideas?

Answer 1

Let us recall the definition of a closed operator. Let $D(K)\subset X$ denote the domain of $K$. We say that $K$ is closed if for every sequence $(x_n)_{n=1}^\infty$ in $D(K)$ that converges to some $x\in X$ such that $Kx_n\to y$ for some $y\in Y$ we have $x\in D(K)$ and $Kx=y$.

Suppose that $(x_n)_{n=1}^\infty$ is a convergent sequence in the kernel of $K$ and let $x\in X$ be its limit. Then $Kx_n=0$ for all $n$, so the sequence $(Kx_n)_{n=1}^\infty$ is convergent with limit $y=0$. By the hypothesis, $Kx=0$, so $x$ is in the kernel of $K$.

Answer 2

The subspace $X \times \{0\}$ is a closed subspace of $X\times Y$. If $\mathcal{G}(K)$ is the graph of $K$ in $X\times Y$ and $\mathcal{N}(K)$ is the null space of $K$, then $$ \mathcal{G}(K) \cap (X\times \{0\}) = \mathcal{N}(K) \times \{0\} $$ is closed, which also makes $\mathcal{N}(K)$ closed.