# Spectral theorem for bounded self-adjoint and coercive operators on a Hilbert space

I think that the title is quite clear. A spectral theorem for bounded self-adjoint operators on Hilbert spaces is quite standard in the literature. What if the operator is also coercive?

• What is your quesrion? – AnyAD Jun 27 '18 at 8:26
• I am wondering if the spectral theorem can be in any way strengthened if one additionally assumes coercivity. – tks Jun 28 '18 at 8:34

## 2 Answers

I will try to say something. The spectral theorem tells us that if $T\in \mathcal{L}(H)$ is self-adjoint, then there is a real-valued function $\varphi\in L^{\infty}(X)$ for some $\sigma$-finite measure space $X$, and a unitary operator $U:H\to L^2(X)$ such that $$T=U^{*}M_{\varphi}U$$ where $M_{\varphi}:L^2(X)\to L^2(X)$ is the multiplication operator associated to $\varphi$. Suppose $T$ is coercive, i.e. for some $c>0$ we have $$\left \langle Tx,x\right\rangle_H \geq c\|x\|_H^2,\qquad \forall x\in H$$ then $$\left \langle Tx,x\right\rangle_H = \left \langle U^*M_{\varphi} Ux,x\right\rangle_H =\left \langle M_{\varphi}Ux,Ux\right\rangle_{L^2} \geq c\|Ux\|_{L^2}^2,\qquad \forall x\in H$$ Since $U$ is invertible, substituting $f=Ux$ the above is equivalent to $$\int_X \varphi |f|^2= \left \langle M_{\varphi}f,f\right\rangle_{L^2}\geq c\|f\|_{L^2}^2=c\int_X |f|^2,\qquad \forall f\in L^2(X)$$ i.e. $$\int_X (\varphi-c)|f|^2\geq 0,\qquad \forall f\in L^2(X)$$ And therefore $\varphi \geq c$ a.e.

In other words, a bounded self-adjoint operator is coercive iff the function $\varphi$ associated by the spectral theorem is bounded from below by the coercivity constant $c$.

• Good point! Is it possible to combine this with the boundedness of the spectrum? (See comment below) – tks Jun 28 '18 at 9:34
• Since $T$ is unitarily equivalent to $M_{\varphi}$, they have the same spectrum, so $\sigma(T)=\sigma(M_\varphi)=\operatorname{ess\,ran}\varphi\subset [c,\|\varphi\|_{\infty}]$. – Lorenzo Quarisa Jun 28 '18 at 10:07
• Why the essential range? Don't we then miss the eigenvalues, if eg $\varphi$ is piecewise constant? Do you have a reference for $\sigma(M_\varphi)=range(\varphi)$? – tks Jun 29 '18 at 9:08
• $\operatorname{ess\,ran}\varphi=\left\{y\in \mathbb{R}: \mu(\varphi^{-1}(y-\varepsilon,y+\varepsilon))>0\;\forall \varepsilon >0\right\}$ (where $\mu$ is the measure on $X$) the eigenvalues correspond to $\sigma_p(M_{\varphi})=\left\{y\in \mathbb{R}:\mu(\varphi^{-1}(y))>0\right\}$. Therefore, clearly $\sigma_p(M_{\varphi})\subset \operatorname{ess\,ran}\varphi$. If $\varphi(x)=y_0$ on a positive measure set, then $\mu(\varphi^{-1}(y_0))>0$ and therefore $y_0$ is both an eigenvalue and an element of the spectrum. – Lorenzo Quarisa Jun 29 '18 at 9:26
• For a reference see Rajendra Bhatia - Notes on functional analysis, Proposition 9 at page 149. By the way, it would not make sense to talk about the range of $\varphi$, since it is a $L^{\infty}$ function which is not even defined pointwise. – Lorenzo Quarisa Jun 29 '18 at 9:31

A bounded selfadjoint operator $A$ satisfies $\inf_{\|x\|=1}\langle Ax,x\rangle = \mu$ iff the spectrum of $A$ is contained in $[\mu,\infty)$. The inf is always the lower bound for the spectrum. The sup is always the upper bound for the spectrum. Both of these numbers, inf and sup, are always in the spectrum.

• Good point! Since the operator is bounded one also gets that $\sigma(A)\subseteq [c_{coerc},c_{cont}]$. – tks Jun 28 '18 at 9:34
• @lks : And the endpoints are included in the spectrum. – DisintegratingByParts Jun 28 '18 at 22:15
• Which implies that $c_{coerc}$ is the minimum of $\varphi$ from the answer above...? – tks Jun 29 '18 at 9:11
• I'm not sure I understand your last question @tks . – DisintegratingByParts Jun 29 '18 at 15:30
• I mean that since $\sigma(T)=\sigma(M_\varphi)=\mathrm{ess\,ran}\,\varphi$ then $\mathrm{ess\, inf}\,\varphi=c_{coerc}$. – tks Jul 2 '18 at 8:14