Let $H_i$ be a $\mathbb C$-Hilbert space and $T$ be a densely-defined linear operator from $H_1$ to $H_2$.
How can we show that if $T$ is injective and $\operatorname{im}T$ is dense, then $T^\ast$ is injective as well? I've read that the reason is that $$\ker T^\ast=(\operatorname{im}T)^\perp=\{0\}\tag1,$$ but I don't get why $(1)$ holds.
I know that for a general densely-defined $T$, $\ker T^\ast=(\operatorname{im}T)^\perp$ and hence $(\ker T^\ast)^\perp=\overline{\operatorname{im}T}$. On the other hand, the identity $\ker T=(\operatorname{im}T^\ast)^\perp$ can only be concluded, when $T$ is closable (since this is equivalent to $T^\ast$ being densely-defined).
So, assuming $\operatorname{im}T$ is dense, the only thing I was able to infer is that $$(\ker T^\ast)^\perp=\overline{\operatorname{im}T}=H_2\tag2.$$ Now if$T^\ast$ would be continuous, then $\ker T^\ast$ would be closed and hence $H_2=\ker T^\ast\oplus(\ker T^\ast)^\perp$, which would immediately yield $\ker T^\ast=\{0\}$ and hence the claim.
Best Answer
Converting my comment to an answer..
Suppose $T^*x = 0$ for some $x\in\operatorname{dom}(T^*)$, then $\langle T^* x, y\rangle 0$ for all $y\in H_1$. This then says that $\langle x, Ty\rangle = 0$ for all $y\in \operatorname{dom}(T)$ by definition of the adjoint operator. Since $T$ is densely-defined and has dense range, there is a sequence $(y_n)_{n=1}^{\infty}\subseteq H_1$ such that $(Ty_n)_{n=1}^{\infty}$ converges to $x$—note that nothing is said about $y_n\to y$ (which would be the closedness condition you were probably thinking of)—and so $\langle x, Ty_n\rangle = 0$ since $\langle x, Ty\rangle = 0$ for all $y\in H_1$ and particularly for our sequence. But since $Ty_n \to x$ and inner products are continuous, we can pass the limit inside to conclude that $\langle x, x\rangle = 0$, i.e. $x = 0$.