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!
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.