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Fix $a$ in ${\mathbb R}^d$ and $b$ in ${\mathbb R}$, consider $x$ in ${\mathbb R}^d$ where $\|x\| \le B$. Is the following function Lipschitz? (and if so, what is a Lipschitz constant?) $$f(x) = (\langle a, x \rangle + b)^2$$

Xi Wu
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1 Answers1

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$Df(x) = 2(\langle a, x \rangle + b) a^T$, and $\|Df(x)\| \le 2(|b|+\|a\| \|x\|) \|a\| \le 2(|b|+B\|a\|) \|a\|$.

Then the mean value theorem shows that $f$ is Lipschitz with rank $2(|b|+B\|a\|) \|a\|$.

copper.hat
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