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My question is this:

If $x\in \Omega=\{x:x\in\mathbb{R}^n, ||x||=1\} $, and it is given that $f: \Omega\to \mathbb{R}^n$ is Lipschitz, is the quantity $g(x) = x^T f(x)$ (or, for that matter any inner product, $g(x)=\langle x, f(x)\rangle$) Lipschitz too?

Are there any standard results to provide conditions for this?

vitamin d
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Nidish Narayanaa
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  • I think this is the composition of two Lipschitz functions and that that is enough to imply that it is itself Lipschitz. – Greg Martin Mar 22 '21 at 19:01
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    Write $g(x) - g(y) = \langle x, f(x)-f(y) \rangle + \langle y, f(x)-f(y) \rangle$ and then estimate each term. – Martin R Mar 22 '21 at 19:04

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