This one has been stumping me for quite some time know...
Given continuous on and differentiable on such that there exists an so that for all , show that there exists a unique point such that .
I can show that if there is a point, then it is unique, but showing the point exists is challenging.
Hint: Pick and let .
I have used repeated applications of the mean value theorem to achieve the following inequality
Unfortunately, this is not Cauchy necessarily, so I am very stuck. Any help would be greatly appreciated!
Am I missing something here? You want to prove f(c)= c so that is an "intermediate value" theorem, not a "mean value" theorem.
You need neither that the function is differentiable nor the bound on the derivative to prove "existence", perhaps to prove "uniqueness".
This is a well known proof:
Suppose f:[a,b]->[a,b] is continuous on [a, b]. If f(a)= a, then we are done. So we can assume f(a)> a. If f(b)= b, we are done. So we can assume f(b)< b. Let h(x)= f(x)- x. Then h(a)= f(a)-a> 0 and h(b)= f(b)- b< 0. But h(x) is also continuous so there must exist c such that h(c)= f(c)- c= 0. Therefore f(c)= c.