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

Question

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?

Answer 1

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 $c$.

Answer 2

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.