How to prove this inequality for operator and function

Question

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.

Answer 1

The term on the left is $$ \|Tf \|^2 = \left\|T \frac{f}{\|f\|} \right\|^2 \|f\|^2 \leq \sup\left\{ \left\|T g \right\| \mid \|g\|=1\right\}^2 \|f\|^2 = \|T\|^2 \|f\|^2 $$

Answer 2

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!!!