Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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!

share|cite|improve this answer
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 $

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.