Let ${\bf T} = (T_1,...,T_d) \in \mathcal{B}(F)^d$, with $\mathcal{B}(F)$ denotes the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

It is true that $$\left(\sum_{|\alpha|=n} \|{\bf T}^\alpha x\|\|{\bf T}^\alpha y\|\right)^2\leq \left(\sum_{|\alpha|=n} \|{\bf T}^\alpha x\|^2\right) \left(\sum_{|\alpha|=n} \|{\bf T}^\alpha y\|^2\right)\;?$$

with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; and ${\bf T}^\alpha:=T_1^{\alpha_1} T_2^{\alpha_2}...T_d^{\alpha_d}$

  • $\begingroup$ Of course. This is not "related to" Cauchy-Schwarz, this is simply a special case of Cauchy-Schwarz! $||T^\alpha x||$ is just a number - Cauchy-Schwarz applies to any two sequences of numbers, whether they happen to be the norm of some vector or not. $\endgroup$ – David C. Ullrich Apr 5 '18 at 15:07
  • $\begingroup$ @DavidC.Ullrich I don't understand how to apply Cauchy Schwartz inequality because I know that $$\left( \sum_j \|A_jx\| \|A_jy\|\right)^2\\ \le\left( \sum_j \|A_jy\|^2 \right) \left( \sum_j \|A_jx\|^2 \right)$$ $\endgroup$ – Schüler Apr 5 '18 at 15:11
  • $\begingroup$ But in this situation the sum is under all multi-indice $\alpha$ with $|\alpha|=n$ $\endgroup$ – Schüler Apr 5 '18 at 15:21
  • $\begingroup$ ??? You really think it makes a difference whether the index is "$j$" or "$\alpha$"? That's incredible. $\endgroup$ – David C. Ullrich Apr 5 '18 at 15:25
  • $\begingroup$ Sorry for my bad. But I'm facing difficulties because $\alpha\in \mathbb{N}^d$ and $j\in \mathbb{N}$ $\endgroup$ – Schüler Apr 5 '18 at 15:31

Evidently we know C-S in this form:

CS1 If $x_j,y_j\ge0$ for $j=1,2\dots,n$ then $\left(\sum_1^n x_jy_j\right)^2 \le\sum_1^nx_j^2\sum_1^ny_j^2$

but we're stuck on this version:

CS2 If $F$ is a finite set and $x_\alpha,y_\alpha\ge0$ for $\alpha\in F$ then $\left(\sum_{\alpha\in F}x_\alpha y_\alpha\right)^2\le\sum_{\alpha\in F}x_\alpha^2\sum_{\alpha\in F}y_\alpha^2$.

It is awesomely obvious that the two are equivalent: To show that CS2 implies CS1, let $F=\{1,2,\dots,n\}$. To show that CS1 implies Cs2: Suppose that $F=\{\alpha_1,\dots,\alpha_n\}$. For any $X_\alpha$ defined for all $\alpha\in F$ we have $$\sum_{\alpha\in F}X_\alpha=\sum_{j=1}^n X_{\alpha_j}.$$(Apply that three times, with $X_\alpha=x_\alpha y_\alpha$, $X_\alpha=x_\alpha^2$ and $X_\alpha=y_\alpha^2$.)


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