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Is there any reason that this generalization of the mean value theorem would fail?

Let A be a subset of Rn that is differentiably connected, and let f : A --> R be continuously differentiable at every point in A. If x and y are two points in A then there exists a point c in A such that

f(y)-f(x) = (grad(f(c)))*(y-x)

It seems like this would work to me, however every generalization I've run into specifies that A must be open, so I assume if it failed it would be on account of that, however I can't figure out why.


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A hint: Consider the auxiliary function $\phi(t):=f(x+t(y-x))$ for $0\leq t\leq 1$. In using this function you make an essential assumption about the two points $x,y\in A$. This assumption is the missing link; it has nothing to do with $A$ being open or what.

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Actually, $A$ does have to be open, because you can only apply the chain rule to interior points of a set. – Ryker Oct 5 '14 at 16:29

The important thing is that you have a $C^1$ curve between your points $x,y \in A$, where the curve lies wholly inside the domain where your function $f$ is differentiable.

Now if you imagine a "dumbbell" like set (say, the unit circles centered on (1,0) and (-1,0) in the plane, it is not clear what you mean when you say that $f$ is differentiable inside this domain; specifically, the nature of $\nabla f$ at the origin becomes an interesting question.

You can finesse the definition there, but the standard definition of the derivative assumes that there is some neighborhood about the point where the function is defined.

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A should be a convex nonempty subset Rn so that there should exist M>0 such that ||Vf(X)||<=M O(t)=f(x+t(y-x)), O is differentiable on R,with differential Vf(x+t(y-x))(y-x) by the mean valeu theorem,there exist a 0< t'<1 such that [O(1)-O(0)]/1=(t')'=f(y)-f(x); f(y)-f(x)=Vf(x+t'(y-x))(y-x) where c=x+t'(y-x).

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Welcome to MSE! It is very helpful to write your questions and answers using MathJax as it greatly improves readability. Regards – Amzoti Jan 22 '13 at 23:20

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