How to prove this inequality for operator and function How to prove this?
$\sum_{k=1}^{\infty}|(Tf)_k|^2\leq ||T||^2||f||^2$
where $T$ is an operator and a function $f$. $(Tf)_k$ is the $k$-th coordinate of Tf.
Should this involve Cauchy-Schwarz or the inner product $< ,>$?
Thanks in advance.
 A: The term on the left is 
$$
     \|Tf \|^2
=    \left\|T \frac{f}{\|f\|} \right\|^2 \|f\|^2
\leq \sup\left\{ \left\|T g \right\| \mid \|g\|=1\right\}^2 \|f\|^2
=    \|T\|^2 \|f\|^2
$$
A: Here it is in a nutshell, sans my typical long-windedness;  note that, to make this one fly, Wilbur, we need to assume we are dealing with a separable Hilbert space, so we can invoke a countable orthonormal basis $\mathbf e_k$.  Also, I am obviously assuming that $(Tf)_k = \langle \mathbf e_k, Tf \rangle$.
1.)  $Tf = \sum_1^\infty \langle \mathbf e_j, Tf \rangle \mathbf e_j; \tag{1}$
2.)  $\langle Tf, Tf \rangle = \langle \sum_1^\infty \langle \mathbf e_j, Tf \rangle \mathbf e_j, \sum_1^\infty \langle  \mathbf e_k, Tf \rangle \mathbf e_k \rangle = \sum_{j, k = 1}^{j, k = \infty} \langle \mathbf e_j, \mathbf e_k \rangle \overline{\langle \mathbf e_j, Tf \rangle} \langle \mathbf e_k, Tf \rangle$
$= \sum_{j, k = 1}^{j, k = \infty} \delta _{jk} \overline{\langle \mathbf e_j, Tf \rangle} \langle \mathbf e_k, Tf \rangle = \sum_1^\infty \overline{\langle \mathbf e_k, Tf \rangle} \langle \mathbf e_k, Tf \rangle = \sum_1^\infty  \vert (Tf)_k \vert^2; \tag{2}$
3.)  Also, $\vert \langle Tf, Tf \rangle \vert \le \Vert Tf \Vert^2$ by Cauchy-Schwarz, and $\Vert Tf \Vert \le \Vert T \Vert \Vert f \Vert$, so $\vert \langle Tf, Tf \rangle \vert  \le \Vert T \Vert^2 \Vert f \Vert^2; \tag{3}$
Finally, combining (2) and (3) we have
4.)  $\sum_1^\infty \vert (Tf)_k \vert^2 \le \Vert T \Vert ^2 \Vert f \Vert^2, \tag{4}$
as was to be shown.  QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
