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I am interested in bounding this $$\langle x, Ay \rangle \leq {\rm ?} $$ in terms of the sums of norms of $x$ and $y$ (special case where matrix $A$ can be seen as an identity matrix)?


Partial attempt

Using Cauchy-Schwarz inequality, then I am not sure \begin{align} \langle x, Ay \rangle &\leq \|x \|_2 \|A\|_2 \|y\|_2 \\ &\overset{?}{\leq} \left( \|x \|_2^2 + \|A\|_2^2 + \|y\|_2^2 \right), \end{align} where $\|A\|_2$ is a spectral norm.


Attempt2 (Considering Jean Marie's answer and comment)

Using Cauchy-Schwarz inequality, then applying AM-GM on the norms of $\| x\|$ and $\| y\|$, that is,

\begin{align} \langle x, Ay \rangle &\leq \|x \|_2 \|A\|_2 \|y\|_2 \\ &\leq \frac{\|A\|_2}{2} \left( \|x \|_2^2 + \|y\|_2^2 \right). \end{align}

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There is an important lack of homogeneity drawback to attempt such inequations with additions :

\begin{align} \langle x, Ay \rangle &\overset{?}{\leq} \left( a\|x \|_2^2 + b\|A\|_2^2 + c\|y\|_2^2 \right), \end{align}

(I have added coefficients $a,b,c$ to make the RHS even more general).

I mean by "lack of homogeneity" the fact that for example,

  • if you replace $x$ by $\lambda x$ and $y$ by $\frac{1}{\lambda} y$, the LHS is unchanged, whereas the RHS is changed, becoming an expression of the form

$$u+v \lambda^2 +\dfrac{w}{\lambda^2}$$

that will be difficult to manage for example because it can be made arbitrarily large.

  • if you replace $x$ by $\lambda x$, $y$ by $\lambda y$, $A$ by $\frac{1}{\lambda^2}A$, the LHS is unchanged, whereas the RHS becomes an expression of the form

$$u\lambda^4 +\dfrac{v}{\lambda^4}$$

etc...

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  • $\begingroup$ Thank you for your comment. I think I see your point. How about this approach and ignore the summation of spectral norm of $A$. $\langle x, Ay \rangle \leq \|x \|_2 \|A\|_2 \|y\|_2 \leq \frac{\|A\|_2}{2} \left( \| x \|_2^2 + \| y \|_2^2 \right)$? $\endgroup$
    – learning
    May 23 '20 at 16:10
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    $\begingroup$ I think it's much more "harmonious", with a coefficient $\frac12$ that could be something else ($\alpha$) ; In particular, grouping everything in the RHS and writing it under a "canonical" form $ a'(|x \|_2 + b \|y\|_2)^2 +c \ge 0 $ will be informative about the good $\alpha$ to be chosen ... $\endgroup$
    – Jean Marie
    May 23 '20 at 16:21
  • $\begingroup$ Thank you so much. $\endgroup$
    – learning
    May 23 '20 at 16:22
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As Jean Marie has already pointed out, finding an upper bound purely in terms of sums of $\|x\|$, $\|y\|$ and $\|A\|$ seems rather difficult.

What you could try however is polarization $$\langle x, Ay \rangle = \frac{1}{4}\left(\|x + Ay \|^2 - \|x - Ay\|^2\right) \leq \frac{1}{4}\|x + Ay\|^2 \leq \frac{1}{4}\left(\|x\| + \|Ay\|\right)^2.$$

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