# 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? Jun 27, 2018 at 8:26
• I am wondering if the spectral theorem can be in any way strengthened if one additionally assumes coercivity.
– tks
Jun 28, 2018 at 8:34

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, 2018 at 9:34
• Since $T$ is unitarily equivalent to $M_{\varphi}$, they have the same spectrum, so $\sigma(T)=\sigma(M_\varphi)=\operatorname{ess\,ran}\varphi\subset [c,\|\varphi\|_{\infty}]$. Jun 28, 2018 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, 2018 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. Jun 29, 2018 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. Jun 29, 2018 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, 2018 at 9:34
• @lks : And the endpoints are included in the spectrum. Jun 28, 2018 at 22:15
• Which implies that $c_{coerc}$ is the minimum of $\varphi$ from the answer above...?
– tks
Jun 29, 2018 at 9:11
• I'm not sure I understand your last question @tks . Jun 29, 2018 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, 2018 at 8:14