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Let $f:\mathbb{R}^n \to \mathbb{R}$ be an $L$-lipschitz function. Now, for all $x \in \mathbb{R}^{n}$, define $$\tilde{f}(x) = \left\{ \begin{array}{cc} \|x\|_2 f\left(\frac{x}{\|x\|_2}\right) & x \neq 0\\ 0 & x = 0 \end{array}\right. $$ where $\|\cdot\|_2$ is the Euclidean norm.

Two questions:

(1) I have seen it cited that the lipschitz constant of $\tilde{f}$ is at most $3L$, but I have not been able to prove this. Is there some simple trick I am missing?

(2) If we require that $f$ also satisfies $af(x) = f(a x)$ for constants $a > 0$, it seems obvious that $\tilde{f}$ is $L$-lipschitz, so what is the disconnect?

Any insight would be helpful here!

WazyMaze
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1 Answers1

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The statement is not generally true. Take for example $f(x) = \cos(x) + 10$ with $L=1$, then: $$ \tilde{f}(x) = \lvert x\rvert(\cos(x/\lvert x\rvert) + 10) = \lvert x\rvert(\cos(1) + 10) , $$ which has smallest Lipschitz constant $\tilde{L} = \cos(1) + 10 > 3$.

However, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ such that $\lVert x\rVert = \lVert y\rVert = a$, then $$ \lvert\tilde{f}(x) - \tilde{f}(y)\rvert=a\lvert f(x/a) - f(y/a)\rvert\leq aL\lVert x/a - y/a\rVert = L\lVert x-y\rVert. $$

Now, let $x,y\in\mathbb{R}^n\setminus\{0\}, x\neq y,$ be arbitrary and assume w.l.o.g. that $a = \lVert x\rVert > \lVert y \rVert$, and let $z = ay/\lVert y \rVert$, $t=\lVert y\rVert/a, 0 < t < 1$, then $$ \lvert\tilde{f}(z) - \tilde{f}(y)\rvert = \lvert af(z/a) - taf(z/a)\rvert = a\lvert f(z/a)\rvert\lvert t-1\rvert \leq a(1-t)(\lvert f(z/a) - f(0)\rvert + \lvert f(0)\rvert)\leq a(1-t)(L + \lvert f(0)\rvert). $$

Therefore $$ \lvert\tilde{f}(x) - \tilde{f}(y)\rvert \leq \lvert\tilde{f}(x) - \tilde{f}(z)\rvert + \lvert\tilde{f}(z) - \tilde{f}(y)\rvert \leq L\lVert x-z\rVert + a(1-t)(L + \lvert f(0)\rvert)\leq\\ L\lVert x-y\rVert + L\lVert y-z\rVert + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + taL\lvert 1/t - 1 \rvert + a(1-t)(L + \lvert f(0)\rvert)=\\ L\lVert x-y\rVert + aL(1-t) + a(1-t)(L + \lvert f(0)\rvert) = L\lVert x-y\rVert + a(1-t)(2L + \lvert f(0)\rvert). $$

Now, $$ \lVert x \rVert = a \leq \lVert x-y\rVert + at \implies \frac{1}{ \lVert x-y\rVert}\leq \frac{1}{a(1-t)}, $$ thus $$ \frac{\lvert\tilde{f}(x) - \tilde{f}(y)\rvert}{\lVert x-y\rVert}\leq L + \frac{a(1-t)(2L + \lvert f(0)\rvert)}{\lVert x-y\rVert} \leq 3L + \lvert f(0)\rvert. $$

Because the choice of $x$ and $y$ is arbitrary, we conclude that the Lipschitz constant of $\tilde{f}$ is no larger than $3L + \lvert f(0)\rvert$.

V.S.e.H.
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