$ker(T)^{bot} = overline{im(T^*)}$ if $T$ is a linear operator between Hilbert spaces

For a Hilbert space, and any vectors $x,y$, $$langle T^*x,yrangle = langle x,Tyrangle$$ Hence $xin(im T)^perpiff xinker T^*$, i.e., $(im T)^perp=ker T^*$. Taking a second perp gives $$(ker T^*)^perp=(im T)^{perpperp}=overline{im T}$$ Apply this identity with $T^*$ instead of $T$ to get $$(ker T)^perp=overline{im T^*}$$

Edit: This works in Hilbert spaces but not Banach spaces because $T^{**}=T$ is valid in the former but not the latter in general.

Edit 2: Direct proof. Let $xinoverline{im(T^*)}$ and $yinker T$, then there are vectors $z_n$ such that $T^*z_nto x$, and $Ty=0$. So $$langle y,xrangle=lim_{ntoinfty}langle y,T^*z_nrangle=lim_{ntoinfty}langle Ty,z_nrangle=0$$