Does a linear map preserving norm also approximately preserve inner products?

Question

Suppose I have distribution on linear map $D_{\epsilon,\delta}$ over $\mathbb{R}^{d\times d}$ such that for any $0<\epsilon,\delta<1/2$ we have $\forall x$, $$\Pr_{A \sim D_{\epsilon,\delta}}(\lvert \lVert Ax\rVert^2- \lVert x\rVert^2\rvert>\epsilon \lVert x\rVert^2)<\delta$$ Now given this can I say that $A$ drawn from $D_{\epsilon,\delta}$ also preserves inner product in the following sense $$\Pr_{A \sim D_{\epsilon,\delta}}(\left\lvert \langle Ax_1,Ax_2\rangle-\langle x_1,x_2\rangle\right\rvert>\epsilon\lVert x_1\rVert\lVert x_2\rVert)<\delta$$ I tried using polarization identity to get the result, I have following thing so far \begin{align*} \langle Ax_1,Ax_2\rangle&=\frac{1}{4}(\lVert A(x_1+x_2)\rVert^2-A(x_1-x_2)\rVert^2) \\ &\le\frac{1}{4}((1+\epsilon)\lVert x_1+x_2\rVert^2-(1-\epsilon)\lVert x_1+x_2\rVert^2)\\ &=\langle x_1,x_2 \rangle+\frac{\epsilon}{2}(\lVert x_1\rVert^2+\lVert x_2\rVert^2) \end{align*} Now I am not able to go further as $\lVert x_1\rVert^2+\lVert x_2\rVert^2\geq 2\lVert x_1\rVert\lVert x_2\rVert$. Any help, comments, hints are greatly appreciated. Thanks.

Answer 1

You have proved that $$|\langle Ax,Ay \rangle - \langle x,y \rangle| \leq \frac{\varepsilon}{2}(\|x\|^2+\|y\|^2) = \varepsilon,$$ for all vectors $x$ and $y$ of length $1$. Hence for all vector of any length : $$|\langle Ax,Ay \rangle - \langle x,y \rangle| \leq \varepsilon.\|x\|.\|y\|.$$