1
$\begingroup$

If we have an orthonormal family, $\{u_n\}_{i=1}^\infty$ in a Hilbert Space $H$, I need to show that for $x\in H$ we have the following inequality:

$\left|\left\{n|\langle x, u_n \rangle > \frac{1}{m}\right\}\right|\leqslant m^2||x||^2.$

The only thing that seems relavent to me is the Bessel inequality $\sum_{n=1}^{\infty} |\langle mx,u_n\rangle |\leq ||mx||^2$ but this leads to trying to show that $\sum_{n=1}^{\infty} |\langle mx,u_n\rangle \geq |\{n||\langle x, u_n \rangle > \frac{1}{m}\}|$ which doesn't really help me.

It would be great if somebody could give me a hint as to how to go about this.

Thanks very much for any help.

1 Answers 1

1

Let $S_m:=\{n\mid, \langle x,u_n\rangle >m^{—1}\}$. Then $S_m\subset \{n\mid, |\langle x,u_n\rangle|^2 >m^{-2}\}$. This gives, by Bessel's inequality, that $\langle x,x\rangle^2\geqslant \sum_{j\in S_m}|\langle x,u_n\rangle|^2\geqslant \sum_{j\in S_m}m^{-2}=|S_m|m^{-2},$ what we wanted.