Question:
What is a best Lipschitz constant of a gradient of $f(x) = x^T A x$, where $A \in \mathbb{R}^{n \times n}$ is symmetric positive definite and $x \in \mathbb{R}^n$?
My attempt:
The gradient of $f(x)$ is: $\nabla f(x) = 2A x$. Now, plugging in this gradient in the Lipschitz definition,
\begin{align} \left\lVert \nabla f (x) - \nabla f (y) \right\rVert \leq L \left\lVert x - y \right\rVert , \forall x , y \in \mathbb{R}^{n}, \end{align}
such that \begin{align} \left\lVert Ax - Ay \right\rVert = \left\lVert A\left(x - y\right) \right\rVert \leq L \left\lVert x - y \right\rVert. \end{align}
Now, if I apply Cauchy-Schwarz inequality to the left hand side, then \begin{align} \left\lVert 2A\left(x - y\right) \right\rVert \leq 2\left\| A \right\| \left\lVert x - y \right\rVert, \end{align} implies that best $L$ is twice a matrix norm of $A$ induced by a vector norm.
Am I correct?