Inequality for a selfadjoint operator on Hilbert space 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.  
 A: 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:
  
  
*
  
*$h \in D(T)$ iff $Uh \in D(M_\phi)$; and 
  
*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!
A: 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 $  
