The typical example of a compact, injective operator on $\ell^2$ is $$ K = \text{diag}(1,1/2,1/3,...).$$
I'm struggeling to understand why this operator has $\text{ker}(K) = \{0\}$ (which it should given it the injectivity), since for the unit vector basis $ \{e_n\}$ with $e_n = (0,0,...,1,0,...)$ we have $$\lim_{n\to\infty} Ke_n = \lim 1/n = 0,$$ but $\|e_n\| = 1$.
What am I missing here? Is the limit element not a proper element of $\ell^2$, or is it because the sequence $\{e_n\}$ is not convergent?
$\forall x:=(x_n)\in\ell^2\quad(Kx)=(x_n/n)$ is the $0$-sequence iff all $x_n$'s are $0,$ i.e. iff $x=0.$ Hence $\ker K=\{0\}.$