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Let $T$ be a (possibly unbounded) selfadjoint nonnegative operator on a Hilbert space $H$ with domain $D$. Assume that $\langle T u, u \rangle \leq c$ for some $c>0$ and some $u\in D$.

I found stated that $\forall d>0$ the inequality $\|1_{[d,\infty)}(T) \ u \|^2\leq \frac{c}{d}$ holds, where $1_{[d,\infty)}(T)$ denotes the spectral projector of $T$ corresponding to the interval $[d,\infty)$.

How can I prove this? I thought to involve the spectral theorem somehow, but I don't know where to start. Thanks for any help.

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I don't understand this sentence: $\langle T u, u \rangle \leq c$ for some $c>0, u\in D$. Is it for all $u$ or some particular $u$? Either way it does not make much sense to me. –  timur Aug 16 '12 at 15:18
    
I mean for a particular $u$: there exist a c>0 and a $u\in D$ such that $\langle Tu,u\rangle\leq c$. And , sorry i wrote $b$ instead of $d$. I'm going to correct it, hoping it's clear now. –  Hans Aug 16 '12 at 15:25
    
Ok I see it makes sense because you used the same $u$ in the inequality. –  timur Aug 16 '12 at 15:27
    
@ByronSchmuland. Thanks for your comment. If $T=dI$ then $T = d 1_{[d,\infty)}(T)$ , and everything works. –  Hans Aug 16 '12 at 17:00

2 Answers 2

up vote 2 down vote accepted

I think it's also instructive to look at this via the multiplication operator version of the spectral theorem. I'm not sure offhand of a canonical reference for this statement (can anyone supply one?) so I'll give it here.

Definition. Let $(X,\mu)$ be a measure space, and $\phi : X \to \mathbb{R}$ a measurable function. The multiplication operator $M_\phi$ corresponding to $\phi$ is the (possibly) unbounded operator on $L^2(X)$ with domain $D(M_\phi) = \{ f \in L^2(X) : \int_X |\phi(x) f(x)|^2\,\mu(dx) < \infty\}$ and defined by $(M_\phi f)(x) = \phi(x) f(x)$.

It is a simple exercise to show that:

  • $M_\phi$ is self-adjoint

  • $M_\phi$ is a bounded operator iff $\phi$ is an (essentially) bounded function

  • $M_\phi$ is nonnegative definite iff $\phi \ge 0$ almost everywhere

It is also easy to see how the functional calculus works for such operators: if $g : \mathbb{R} \to \mathbb{R}$ is measurable, then $g(M_\phi) = M_{g \circ \phi}$.

Theorem (Spectral Theorem). Suppose $T$ is a self-adjoint operator on a Hilbert space $H$. There exists a measure space $(X,\mu)$, a measurable function $\phi : X \to \mathbb{R}$, and a unitary operator $U : H \to L^2(X)$ such that:

  1. $h \in D(T)$ iff $Uh \in D(M_\phi)$; and

  2. For all $h \in D(T)$, $Th = U^{-1} M_\phi U h$ (i.e. $T = U^{-1} M_\phi U$).

Informally, this says that, up to unitary equivalence, any self-adjoint operator is a multiplication operator. This also gives a functional calculus: $g(T) = U^{-1} M_{g \circ \phi} U$.

Now back to your problem. Thanks to this version of the spectral theorem, it suffices to handle the case when $T$ is a multiplication operator $M_\phi$ on some measure space $(X,\mu)$. Since $T$ is nonnegative, $\phi \ge 0$ a.e. Now we are trying to estimate $$\|1_{[d,\infty)}(T) u\|^2 = \int_X |1_{[d,\infty)}(\phi(x)) u(x)|^2\,\mu(dx) = \int_X 1_{\{\phi \ge d\}}\,u^2\,d\mu = \nu(\{\phi \ge d\})$$ where $\nu$ is the measure $\nu(B) = \int_B u^2\,d\mu$.

On the other hand, $$\langle Tu , u \rangle = \int_X \phi u^2\,d\mu = \int_X \phi \,d\nu.$$

So we can rewrite the statement we want to prove as: $$\nu(\{\phi \ge d\}) \le \frac{1}{d} \int_X \phi\,d\nu.$$ But this is nothing but Markov's inequality!

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I like this point of view very much. Thanks. I think a canonical reference for the spectral theorem you cite could be Theorem VIII.4 in the first Volume of Reed-Simon. –  Hans Aug 18 '12 at 0:17

I think I got it now. It should follow from the spectral theorem as follows. I use $b$ instead of $d$ in order to not make confusion with the $d$ of the integral:

$\|1_{[b,\infty)}(T) u\|^2 \ = \ \langle 1_{[b,\infty)}(T) u,u \rangle \ = \int_{b}^\infty d\langle 1_\lambda(T) u,u\rangle \leq \ \int_b^\infty \frac \lambda b \ d\langle 1_\lambda(T) u,u\rangle \ \leq \ \int_0^\infty \frac \lambda b \ d\langle 1_\lambda(T) u,u\rangle \ = \ \frac 1b \langle Tu,u\rangle \ \leq \ \frac cb $

EDIT. clarification for the first equality: using that $1_{[b,\infty)}(T)$ is selfadjoint (becuase function of a selfadjoint operator) and that being a projection $[1_{[b,\infty)}(T)]^2 = 1_{[b,\infty)}(T)$ we have

$\|1_{[b,\infty)}(T) u\|^2 \ = \ \langle 1_{[b,\infty)}(T) u, 1_{[b,\infty)}(T) u \rangle \ = \ \langle [1_{[b,\infty)}(T)]^2 u, u \rangle \ = \ \langle 1_{[b,\infty)}(T) u,u \rangle $

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I use that $1_{[b,\infty)}(T)$ is selfadjoint and that it is equal to its square. Therefore both the things you wrote are correct in my opinion. –  Hans Aug 16 '12 at 17:14
    
Oh I see now. I will delete my comments. –  Byron Schmuland Aug 16 '12 at 17:15
    
Why $\|1_{[b,\infty)}(T) u\|^2 \ = \ \langle 1_{[b,\infty)}(T) u,u \rangle$? –  timur Aug 16 '12 at 18:00
    
@timur: I edited the answer to clarify. –  Hans Aug 16 '12 at 18:33

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