# Lipschitz contradiction

Assume that $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ is a smooth vector field, and assume that we can find vectors $y_k,x_k$ ($k$ positive integer) such that $(\phi(x_k)-\phi(y_k),x_k-y_k)\geq k \mid y_k-x_k\mid^2$, where $(,)$ is the usual scalar product. Why does this condition contradict the fact that $\phi$ is Lipschitz?

-
Hint: Cauchy-Schwarz. Caution: if you do not assume that $y_k\ne x_k$, your condition becomes empty. – Did Apr 18 '11 at 17:43

## 1 Answer

Recall Cauchy Schwarz which says $(x,y)\leq \|x\| \|y\|$.

Then $$(\phi(x_k)-\phi(y_k),x_k-y_k)\leq \|\phi(x_k)-\phi(y_k)\| \|x_k-y_k\|$$ Since $\phi$ is Lipschitz, there exists a constant $C$ such that $$|\phi(x_k)-\phi(y_k)\|\leq C\|x_k-y_k\|$$ and hence $$(\phi(x_k)-\phi(y_k),x_k-y_k)\leq C|x_k-y_k\|^2.$$ But if we had two sequences, $x_k$, $y_k$ with $$(\phi(x_k)-\phi(y_k),x_k-y_k)\geq k\|y_k-x_k\|^2,$$ we would have to have $C>k$ for every integer $k$ which is impossible.

Hope that helps,

Note: I assume that your statement is meant to be interpreted as "for every integer $k$ we can find $x_k$, $y_k$ ..." rather than "there exists $k$ with..." since setting $\phi$ equal to $k$ times the identity deals with the second case.

-
Did you read the Caution part of my comment? The reason why your (accepted) proof is (technically) false is instructive. – Did Apr 20 '11 at 6:22