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kingwinner
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Q: Prove that if f: R^n -> R is defined by f(x)=arctan(||x||), then |f(x)-f(y)| <= ||x-y|| FOR ALL x,y E R^n.
[<= means less than or equal to]
Theorem: (a corollary to the mean value theorem)
Suppose f is differentiable on an open, convex set S and ||gradient [f(x)]|| <= M for all x E S. Then |f(b) - f(a)| <= M ||b-a|| for all a,b E S.
Now the trouble is that f(x)=arctan(||x||) is not differentiable at x=0 E R^n since the partial derivatives doesn't exist at x=0. Even worse, notice that S = R^n \ {0} is not convex.
Then how I can show that "f is differentiable on an open, convex set S"? (what is S in this case?) I strongly believe that this is an important step because if the conditions in the theorem are not fully satisifed, then there is no guarantee that the conclusion will hold. But this seems to be the only theorem that will help. What should I do?
Next, how can I prove that the conclusion is true FOR ALL x,y E R^n ?
Thanks for explaining!
[<= means less than or equal to]
Theorem: (a corollary to the mean value theorem)
Suppose f is differentiable on an open, convex set S and ||gradient [f(x)]|| <= M for all x E S. Then |f(b) - f(a)| <= M ||b-a|| for all a,b E S.
Now the trouble is that f(x)=arctan(||x||) is not differentiable at x=0 E R^n since the partial derivatives doesn't exist at x=0. Even worse, notice that S = R^n \ {0} is not convex.
Then how I can show that "f is differentiable on an open, convex set S"? (what is S in this case?) I strongly believe that this is an important step because if the conditions in the theorem are not fully satisifed, then there is no guarantee that the conclusion will hold. But this seems to be the only theorem that will help. What should I do?
Next, how can I prove that the conclusion is true FOR ALL x,y E R^n ?
Thanks for explaining!
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