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Background

I am aware of the following:

For a quadratic function $f$ on $\mathbb{R}^2$ defined as $$f(x) = \frac{1}{2}x^TQx - b^Tx,$$ if $Q$ is positive semidefinite, then $L = \lambda_{\max}$.

Question

This result cannot be used in the case we have a quadratic function in the same form, but $Q$ is not positive semidefinite. This is because $Q$'s symmetry is necessary in the proof of the above claim (which I can post as an edit if necessary).

Moreover, from my brief reading of previous questions on this website slightly related to this one, it seems that there may not always be a suitable Lipschitz constant on the domain of $f$ for a quadratic function.

To make things more concrete, let us focus on the example:

\begin{equation*} x^T \begin{pmatrix} 4 & 2\sqrt{2}\\ 0 & 5 \end{pmatrix} x + x^T \begin{pmatrix} 3\\ 6 \end{pmatrix} \end{equation*}

Note that the "$Q$" here has two distinct eigenvalues, 10 and 8. A direction I am considering going in is using the fact that the Hessian is independent of $x$ for all $x \in \mathbb{R}^2$. Admittedly I am not very sure how to proceed from such an observation.

Therefore, my question is:

Is there some way to find a Lipschitz constant for a function of the described nature?

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    $\begingroup$ Note that you can always rewrite a quadratic function in terms of a symmetric matrix $A=(Q+Q')/2$. $\endgroup$ – Brian Borchers Oct 1 '18 at 12:10

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