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Question:

let

$$D=\{u=(x,y)\in R^2\colon||u||=\sqrt{x^2+y^2}\le\dfrac{1}{2}\}$$ and $f(u)=f(x,y)$ is all plane continuously differentiable,and such $$||\nabla f(0,0)||=1,||\nabla f(u)-\nabla f(v)||\le||u-v||$$

let $\forall u,v\in D$, show that:

the function $f(x,y)$ have only points to obtain the maximum value.

My try: since $$||\nabla f(u)-\nabla f(v)||\le||u-v||$$ so I want use Lipschitz continuity,But at last,

I can't work.

and this problem is from this :http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=91&Id=12243&page=9

Thank you for you help!

math110
  • 93,304

1 Answers1

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I think that problem must be modified :

$f$ cannot have a maximum in interior

Note that $$ 1=\|\nabla f((0,0)) \|\leq \| \nabla f((0,0)) - \nabla f({\bf x}) \| + \| \nabla f({\bf x}) \| $$

If ${\bf x}$ is an interior point, then $$\| \nabla f((0,0)) - \nabla f({\bf x}) \|<\frac{1}{2} $$ so that $$ \frac{1}{2}\leq\| \nabla f({\bf x}) \| $$

That is, if ${\bf x}$ is a maximum point, then $ \nabla f({\bf x}) =0$. So, it is a contradiction.

HK Lee
  • 19,964