Dense range if a linear self-adjoint is injective. I'm solving some exercises of Kreyszig's Functional Analysis about self-adjoint operators.
If a self-adjoint linear operator $T:\mathcal{D}(T)\rightarrow H$ is injective, show that
(a) $\overline{\mathcal{R}(T)}=H$ (b) $T^{-1}$  is self-adjoint.
I'd like to prove that $\mathcal{R}(T)^{\bot}=\{0\}$ to finish a), but I have troubles proving it. I tried to use the fact that $T$ is inyective gives that $Kern(T)=\{0\},$ but I get useless things.
For part b), part a) gives the existence of inverse adjoint, but I'm stuck proving that is self-adjoint.
Any suggestion?
 A: For a densely defined operator $A$ we have, by definition,
$$
D(A^*) = \{y\in H\,|\,\exists z_y\in H\,\forall x\in D(A) : (Ax,y) = (x,z_y)\}.
$$
And if $y\in D(A^*)$, then $A^*y = z_y$. From here you easily see that $R(A)^\perp = N(A^*)$, where $R$ and $N$ denote range and kernel, respectively. Here's how it goes: $y\in N(A^*)$ means $A^*y = 0$, i.e., $y\in D(A^*)$ and $z_y = 0$. This is equivalent to $(Ax,y) = 0$ for all $x\in D(A)$. That is, $y\in R(A)^\perp$.
Now, let $T$ be selfadjoint such that $N(T) = \{0\}$. Then $R(T)^\perp = N(T^*) = N(T) = \{0\}$. Thus, $\overline{R(T)} = H$. Hence, $T^{-1}$ is densely defined (on $R(T)$). We have $T^{-1} : R(T)\to D(T)$. Now,
\begin{align*}
D((T^{-1})^*)
&=\left\{x\in H\,|\,\exists z_x\in H\,\forall y\in R(T) : (T^{-1}y,x) = (y,z_x)\right\}\\
&= \left\{x\in H\,|\,\exists z_x\in H\,\forall u\in D(T) : (T^{-1}Tu,x) = (Tu,z_x)\right\}\\
&=\left\{x\in H\,|\,\exists z_x\in H\,\forall u\in D(T) : (u,x) = (Tu,z_x)\right\}.
\end{align*}
But the latter means exactly that $z_x\in D(T^*) = D(T)$ and $Tz_x = x$. This is equivalent to $x\in R(T) = D(T^{-1})$. Thus, $D((T^{-1})^*) = D(T^{-1})$. Let $x$ be a vector in this set. Then from the first line above you see that $(T^{-1})^*x = z_x$. But we also saw that $Tz_x = x$, i.e., $z_x = T^{-1}x$. This finally implies $(T^{-1})^*x = z_x = T^{-1}x$ and therefore $(T^{-1})^* = T^{-1}$. That is, $T^{-1}$ is selfadjoint.
