Currently I'm self studying functional analysis, namely self-adjoint operators. In the text, the author gives the following exercise:
Exercise A: If $T$ is a self-adjoint operator, then $\text{ker}T\perp\overline{\text{Im}T}$.
I think I've tackled this on my own, but I can't find this problem anywhere to verify my solution below. Hence, I'm asking here: is this solution correct?
Solution: Let $x\in\text{ker}T$ and $y\in\overline{\text{Im}T}$. By definition it follows that $Tx=0$ and $\exists\left(Tx_n\right)\in\text{Im}T$ for which $Tx_n\to y$. Therefore by the continuity of the inner product and self-ajointness of $T$, $$ \langle x,y \rangle= \lim\langle x,Tx_n \rangle= \lim\langle Tx,x_n \rangle= \lim\langle 0,x_n \rangle= \lim0\cdot\langle 0,x_n \rangle= 0. $$ This concludes the proof.
EDIT:
I actually misread the exercise. The exercise is actually the following:
Exercise B: If $T$ is a self-adjoint operator, then $\text{ker}T\perp\text{Im}T$.
But this is simpler. Take $x\in\text{ker}T$ and $Ty\in\text{Im}T$. Then by the self-ajointness of $T$, $$ \langle x,Ty \rangle= \langle Tx,y \rangle= \langle 0,y \rangle= 0\cdot\langle 0,y \rangle= 0. $$ This concludes the proof.
