# How to upper bound this $\langle x, Ay \rangle \leq {\rm ?}$ in terms of the sum of norms of $x$ and $y$?

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}

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...

• Thank you for your comment. I think I see your point. How about this approach and ignore the summation of spectral norm of $A$. $\langle x, Ay \rangle \leq \|x \|_2 \|A\|_2 \|y\|_2 \leq \frac{\|A\|_2}{2} \left( \| x \|_2^2 + \| y \|_2^2 \right)$? May 23 '20 at 16:10
• 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 ... May 23 '20 at 16:21
• Thank you so much. May 23 '20 at 16:22

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 $$\langle x, Ay \rangle = \frac{1}{4}\left(\|x + Ay \|^2 - \|x - Ay\|^2\right) \leq \frac{1}{4}\|x + Ay\|^2 \leq \frac{1}{4}\left(\|x\| + \|Ay\|\right)^2.$$