Spectral calculus and inverse

Question

Let $T:\mathcal{D}(T)\to\mathcal{H}$ be a self-adjoint and unbounded operator such that $0\notin\sigma(T)$. As a consequence, $T$ is invertible with bounded inverse. Let now $f\in C(\sigma(T),\mathbb{C})$ be a (not necessarily bounded) continuous function such that $f\neq 0$ point-wise. I was wondering whether the following holds:

$$f(T)=(1/f)(T^{-1})$$

where $1/f$ denotes the (algebraic) inverse, i.e. $(1/f)(\lambda):=1/f(\lambda)$ and where both operators are defined w.r.t. to the spectral theorem of $T$ and $T^{-1}$, respectively. Spelled out, we have to show that

$$\int_{\sigma(T)}\lambda\mathrm{d}\mu^{T}(\lambda)=\int_{\sigma(T^{-1})}\lambda^{-1}\mathrm{d}\mu^{T^{-1}}(\lambda)$$

where $\mu^{A}$ denotes the spectral measure of an operator $A$. Since $\sigma(T^{-1})=\{\lambda^{-1}\mid\lambda\in\sigma(T)\}$, this is equivalent to show that $\mu^{T}(\{\lambda\})=\mu^{T^{-1}}(\{\lambda^{-1}\})$. Alternatively, by the spectral theorem, it should be true that $(1/f)(T^{-1})=f(T^{-1})^{-1}$, so we want to show equivalently that the inverse of $f(T)$ is given by $f(T^{-1})$.

Answer 1

There might be a more direct way, but I will show the following identity using the uniqueness of the functional calculus for $T$ on the class $\mathcal{C}_0$ of continuous functions on $\mathbb{R}$ vanishing at infinity: $$\forall f \in \mathcal{C}_0,\quad f(T) = \left(x \mapsto f\left(\frac1x\right)\right)\left(T^{-1}\right)$$ noting that the function $g_f : x \mapsto f\left(\frac1x\right)$ is well-defined as it remains continuous at $0$ with $g_f(0) = 0$, and $g_f$ is bounded.
You can probably extend this to other measurable functions but you'd have to make sure you find a well-defined analogue for the right-hand side.

1. $f \mapsto g_f\left(T^{-1}\right)$ is an algebra homomorphism because the Borel functional calculus for $T^{-1}$ is one and $f \mapsto g_f$ is too;
2. We have $g_{\overline{f}} = \overline{g_f}$, hence $g_f\left(T^{-1}\right)^* = g_{\overline{f}}\left(T^{-1}\right)$;
3. We can see that $\left\|g_f\left(T^{-1}\right)\right\| \leqslant {\left\|g_f\right\|}_{\infty} = {\|f\|}_\infty$;
4. Let's denote by $r_z$ the function $x \mapsto \frac{1}{z - x}$ for $z \in \mathbb{C} \setminus \mathbb{R}$. Then, we get: $$\begin{split}g_{r_z}\left(T^{-1}\right) &= \left(x \mapsto \frac{1}{z - \frac{1}{x}}\right)\left(T^{-1}\right)\\ &= \left(x \mapsto -\frac{x}{z} \cdot \frac{1}{\frac{1}{z} - x}\right)\left(T^{-1}\right)\\ &= -\frac1z T^{-1} \circ \left(\frac{1}{z}\operatorname{Id} - T^{-1}\right)^{-1}\\ &= \left(\left(\frac{1}{z}\operatorname{Id} - T^{-1}\right) \circ (-zT)\right)^{-1}\\ &= \left(z\operatorname{Id} - T\right)^{-1}\end{split}$$
5. Let $f$ be such that its support is disjoint from $\sigma(T)$. We cannot directly conclude that the support of $g_f$ is disjoint from $\sigma\left(T^{-1}\right)$ as $0$ might or might not belong to any or both of these sets, however we do have that they are disjoint on $\mathbb{R} \setminus \{0\}$. As $g_f(0) = 0$, it is possible to conclude that $g_f\left(T^{-1}\right) = 0$, since $g_f$ is equal to $0$ on $\sigma\left(T^{-1}\right)$ no matter what.

$f \mapsto g_f\left(T^{-1}\right)$ thus verifies the five properties which make the $\mathcal{C}_0$ functional calculus unique, hence it is said functional calculus, and we are done.