# From continuous to bounded Borel functions

I know that we can extend the functional calculus of bounded self-adjoint operators to bounded Borel functions. I want to do the same for unbounded self-adjouint operators.

Therefore assume that $T$ is a unbounded self-adjoint operator on a Hilbert space $H$. Assume that one know how to obtain $f(T)$ if $f$ is continuous on the spectrum of $T$. I want to extend this to Borel functions on the spectrum. Can this be done?

• The usual way is the spectral theorem ... $T$ is similar to a "multiplication" operator on some $L^2$ space. Of course if $T$ is an unbounded operator, then the multiplier is an unbounded function. – GEdgar Jan 13 '15 at 18:29
• @GEdgar: Hence there is an isometric isomorphism from $H$ to $L^2(M,\mu)$ such that the image of $T$ under this isomorphism is a multiplication operator? But how to conclude the borel calculus from this? – Nicky Ihsan Jan 13 '15 at 18:34
• @NickyIhsan: How do you obtain the continuous calculus for unbounded operators?? You can induce a bounded operator by either the Cayley transform or by the positive square root. Then this gives rise to a spectral measure which in turn gives back the desired spectral measure. (The details are as always much more subtle, of course.) – C-Star-W-Star Jan 13 '15 at 20:48

A spectral measure $E$ comes out of the classical version of the Spectrum Theorem. The only difference between $E$ for a bounded selfadjoint operator and an unbounded one is that the support of $E$ for a bounded selfadjoint operator is bounded. For any Borel function $f$, you can define $$f(T)x = \int_{-\infty}^{\infty} f(\lambda)dE(\lambda)x$$ on the domain $\mathcal{D}(f(T))$ consisting of all $x\in X$ for which $$\|f(T)x\|^{2} = \int_{-\infty}^{\infty} |f(\lambda)|^{2}d\|E(\lambda)x\|^{2} < \infty.$$ That turns out to define a closed, densely-defined normal operator whose adjoint $f^{\star}(T)=f(T)^{\star}$ has the same domain at $f(T)$. If $f$ is bounded on $\sigma(T)$, then $f(T) \in \mathcal{L}(H)$.

If $f$ is Borel measurable and $g$ is a bounded Borel function, then $$g(T) : \mathcal{D}(f(T))\subseteq \mathcal{D}(f(T)),$$ and $g(T)f(T)=f(T)g(T)=(fg)(T)$ on $\mathcal{D}(f(T))$; even though the domains of $(fg)(T)$ and $f(T)$ may be different, clearly $\mathcal{D}(f(T))\subseteq\mathcal{D}((fg)(T))$ follows from the above integral characterization of the domain.

Of course, if $f$ and $g$ are bounded Borel functions, then $f(T)$, $g(T)$ and $(fg)(T)$ are defined everywhere and bounded; furthermore, $f(T)g(T)=g(T)f(T)=(fg)(T)$.

Construction of Extended Calculus: A fundamental identity for the functional calculus is the resolvent $$(T-\mu I)^{-1} = \int_{-\infty}^{\infty}\frac{1}{\lambda-\mu}dE(\lambda),\;\;\; \mu \in \rho(T).$$ This solidly (i.e., uniquely) connects $T$ with the spectral measure $E$. These bounded continuous functions $f_{\mu}(t)=(t-\mu)^{-1}$ applied to $T$ are really all that you need to construct the full Borel functional calculus using strong limits. These bounded continuous functions of $T$ give you the resolvent. First, you can construct the spectral measure on intervals: $$\frac{1}{2}\{ E[a,b]+E(a,b)\}x=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}\{f_{u+i\epsilon}(T)-f_{u-i\epsilon}(T)\}x\,du.$$ This is Stone's Formula. This gives you $E[a,b]x$ and $E(a,b)x$ separately as vector limits from outside and inside the interval, respectively. Alternatively, you can form compactly supported continuous functions of $T$ which approach characeristic functions of intervals.

Reference For Stone's Formula: Proving Stone's Formula for Constructively obtaining the Spectral Measure for $A=A^\star$

• Isn't the problem rather the converse: How to obtain a spectral measure from selfadjoint operator than induce a selfadjoint operator by a spectral measure? (Your first sentence gives a hint but afaik it is little harder.) – C-Star-W-Star Jan 13 '15 at 20:42
• @Freeze_S : The statement of the problem: "I know that we can extend the functional calculus of bounded self-adjoint operators to bounded Borel functions. I want to do the same for unbounded self-adjouint operators." I have described that extension based on the Spectral Theorem. – DisintegratingByParts Jan 13 '15 at 20:45
• But how do you obtain a measure from continuous calculus? ("Assume that one know(s) how to obtain $f(T)$ if $f$ is continuous on the spectrum of $T$.") – C-Star-W-Star Jan 13 '15 at 20:54
• @Freeze_S : I've added more. – DisintegratingByParts Jan 13 '15 at 21:21
• That is an interesting approach: So one can circumvent Riesz-Markov & Co by exploiting Stone's formula which gives rise to generating sets for the Borel algebra - nice! (My apologies, I'm not a fan of this, though.) – C-Star-W-Star Jan 14 '15 at 0:22

If $T$ actually is a multiplication operator on $L^2(M,\mu)$, say there is a real-valued function $m$ on $M$ so that $$T(h) = mh$$ for $h \in L^2$, then the functional calculus tells you that $f(T)$ is the multiplication by $f(m)$. The general self-adjoint operator is similar to such a multiplication operator.