Closed linear operators have closed kernels 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? 

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 $K$.
