Bound $\left\lVert Ax \right\rVert$ in term of $\left\lVert x - y \right\rVert$

Question

Let $A \in \mathbb{R}^{n \times n}$ is a symmetric matrix. Given two vectors $x, y \in \mathbb{R}^{n}$ such that $$ \left\langle Ax , x - y \right\rangle < 0 \qquad \textrm{ and } \qquad \left\langle Ax , x - y \right\rangle + c \left\lVert x - y \right\rVert ^{2} = 0 $$ for some $c>0$.

Find a constant $b > 0$ such that $$ \left\lVert Ax \right\rVert \leq b \left\lVert x - y \right\rVert . $$

If the estimate were the other side around then it would be easy since we can apply Cauchy - Schwarz inequality to deduce the bound. However, in this case, it not clear how to proceed. I tried to write the inner product in term of norms \begin{align} \left\langle Ax , x - y \right\rangle & = \dfrac{1}{2} \left\lVert Ax \right\rVert ^{2} + \dfrac{1}{2} \left\lVert x - y \right\rVert ^{2} - \dfrac{1}{2} \left\lVert \left( A - \mathbb{I} \right) x + y \right\rVert ^{2} \\ & = - \dfrac{1}{2} \left\lVert Ax \right\rVert ^{2} - \dfrac{1}{2} \left\lVert x - y \right\rVert ^{2} + \dfrac{1}{2} \left\lVert \left( A + \mathbb{I} \right) x - y \right\rVert ^{2} \end{align} but none of them seem to be useful. Any idea would be appriciated.

Answer 1

In general, there can be no such $b$ even for $n=2$ and $A=I$. Indeed, put $x=(-c,n)$ and $x-y=(1,0)$, where $n$ can be arbitrary big. Then $\|x-y\|=1$, $\langle Ax,x-y\rangle+c\|x-y\|=0$, but $\|Ax\|=\sqrt{n^2+c^2}\|x-y\|$.