Show this function is Lipschitz and find $\|f\|_{Lip}$

Question

Fix a vector $v\in{\mathbb{R}^n}$ and let $f:\mathbb{R}^n\rightarrow{\mathbb{R}}$ be a function defined by $f(x)=<x,v>$ (with $<.,.>$ the euclidean scalar product). Show that $f$ is Lipschitz, with $\|f\|_{Lip}=\|v\|_2$. I tried playing with $|f(x)-f(y)|$ for some $x,y$ and bounding it and pulling out a factor of $\|v\|_2$, but that didn't seem to work.

Any hints?

Answer 1

Recalling that the scalar product is linear on every variable, you have $$|f(x)-f(y)|=|\langle x ; v \rangle - \langle y ; v \rangle|= |\langle x -y; v \rangle| \le ||v||_2 \cdot ||x-y||_2$$ where the inequality is simply Cauchy-Schwarz.