# Lipschitz constant of rescaled function

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!

• Shouldn't the function be defined as $$\tilde{f}(x) = \begin{cases}\lVert x\rVert_2 f(x/\lVert x\rVert_2) & x\neq 0\\0 & x = 0\end{cases}$$? Feb 25, 2021 at 21:36
• Yes, sorry. I'll update. Feb 25, 2021 at 21:40

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 $$3L + \lvert f(0)\rvert$$.