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

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.

## 1 Answer

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 $$\|Ax\|=\sqrt{n^2+c^2}\|x-y\|$$.