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

inequalitylinear algebrareal-analysis

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}

Best Answer

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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