How to show that L2 norm is Lipschitz if the input domain is bounded?

Question

I guess that $||w||_2^2$ is 2B Lipschitz when $||w||<B$. That's because the gradient $||w||_2^2$ is $2w$ and $w$ is bounded, but I can't show my claim according to the definition of the Lipschitzness. Can anybody help me formally prove my claim?

Answer 1

If you define the norm of an element v, as $\left\|v \right\|=sup\frac{<v,x>}{\left\|x \right\|}$ then the Lipschitz continuity follows from the inequality sup(f(x)+g(x))$\leq $supf(x)+supg(x), which implies (set f+g=h) , suph(x)-supf(x) $\leq$ sup(h(x)-f(x)). Since $sup\frac{<v,x>}{\left\|x \right\|}-sup\frac{<w,x>}{\left\|x \right\|}\leq sup\frac{<v-w,x>}{\left\| x\right\|}=\left\|v-w \right\|$. Likewise the l.h.s is $\geq -\left\|v-w \right\|$ and hence $\left|\left\|v \right\|-\left\| w\right\| \right|\leq \left\|v-w \right\|$ (1). Thus, with the standard definition of a norm of an element of the dual space, Lipschitz continuity always holds.In the case of a finite dimensional space and a Euclidean norm, the definition $\left\|v \right\|=\sqrt{v_{1}^{2}+...v_{n}^{2}}$ coincides with $\left\|v \right\|=sup\frac{<v,x>}{\left\|x \right\|}$ and again the L-continuity holds! If you are interested in L-continuity of the square of the norm then by (1) above, $\left|\left\|v \right\|^{2}-\left\| w\right\|^{2} \right|\leq \left\|v-w \right\|\left|\left\| v\right\|+\left\|w \right\| \right|$ which implies that the square of the norm is L-continuous on bounded sets.