0

Let $f: \mathbb{R}^n \to \mathbb{R}^n$ be a $L$-Lipschitz function: $\|f(x)-f(y)\| \le L \|x-y\|$, $L >0$.

How to prove $\|f(x)-f(y)\|^2 \le L \left( f(x)-f(y)\right)^T (x-y)$?

Update: Sorry, I forget to specify an important condition: in fact, $f$ is the gradient of a convex function.

Mengda
  • 175
  • What norm are you working with? And what have you tried thus far? – Keen-ameteur Apr 01 '20 at 14:11
  • 1
    I doubt that such an estimate exists generally. Try $f(x) = -x$ for example. – Martin R Apr 01 '20 at 14:13
  • It is euclidean norm. I can only prove $∥f(x)−f(y)∥^2≤L ∥f(x)−f(y)∥ ∥x−y∥$. I don't know how to pass ∥f(x)−f(y)∥ ∥x−y∥ to (f(x)−f(y))^T(x−y). @Keen-ameteur – Mengda Apr 01 '20 at 14:14
  • Ah That's true... @MartinR Maybe my professor used something else to get this inequality... – Mengda Apr 01 '20 at 14:16
  • 1
    Another counterexample: A rotation by 90 degrees in $\Bbb R^2$: the right-hand side is zero for all $x, y$. – Martin R Apr 01 '20 at 14:18
  • Sorry, I forget to specify an important condition: in fact, $f$ is the gradient of a convex function. @MartinR – Mengda Apr 01 '20 at 14:26
  • 2
    This may be what you are looking for: https://math.stackexchange.com/a/2021718/42969. – Martin R Apr 01 '20 at 14:32

0 Answers0