I have the following proof about unbounded operators but feel a little suspicious, and I would appreciate some help critiquing/verifying it.
The setting is as follows. Suppose $D$ is a first-order self-adjoint elliptic differential operator on a closed manifold $M$ acting on a complex vector bundle $S$ (I have in mind a Dirac operator). One can take the functional calculus of $D$ with respect to the continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ given by $x\mapsto\frac{x}{\sqrt{x^2+1}}$. Denote this operator by
$$f(D):L^2(S)\rightarrow L^2(S),$$
which is now bounded. Let $H^i=H^i(S)$ denote the $i$-th Sobolev space, so that $H^0 = L^2(S)$.
I want to argue that the kernel of $D$ coincides with the kernel of $f(D)$, as follows. Firstly view $D^2+1$ as a closed, self-adjoint, densely-defined operator on $H^1$: $$D^2+1: \text{dom}(D^2+1)=H^2\rightarrow H^0.$$ Since $D$ is self-adjoint, $D^2+1$ has an inverse $H^1\rightarrow H^3$, hence $0$ is not in its spectrum. Use functional calculus to form the following densely-defined operator on $H^0$ $$\sqrt{D^2+1}:\text{dom}\left(\sqrt{D^2+1}\right)\rightarrow H^0,$$ where I believe $\text{dom}\left(\sqrt{D^2+1}\right) = H^1$. This operator is still has an inverse, since $0$ is outside of its spectrum by functional calculus, $$\left(\sqrt{D^2+1}\right)^{-1}: H^0\rightarrow H^1.$$ By the fact that the functional calculus is a homomorphism, we have a well-defined operator $$\left(\sqrt{D^2+1}\right)^{-1} = \sqrt{(D^2+1)^{-1}},$$ and $f(D)$ is the same as the composition $$D\left(\sqrt{D^2+1}\right)^{-1}:H^0\rightarrow H^1\rightarrow H^0.$$ Note that $D(\sqrt{D^2+1})^{-1}$ is elliptic, hence by elliptic regularity every element of its kernel is smooth and so in $H^1$. Then I claim that $D\left(\sqrt{D^2+1}\right)^{-1}$ coincides with the continuous extension of $\left(\sqrt{D^2+1}\right)^{-1}D$ from $H^1$ to all of $H^0$. Given this, it follows that $$\text{ker}(D) = \text{ker}\left(D\left(\sqrt{D^2+1}\right)^{-1}\right) = \text{ker}(f(D)).$$