Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1 / 10E (Analysis I)

The mean value theorem states that a function continuous on $[p,q]$ and differentiable on $(p,q)$ has some $c\in (p,q)$ with $f(q)-f(p)=f^{\prime }(c)(q-p)$.

Take any $p<q$ inside the given open interval. Differentiability implies the continuity needed on $[p,q]$, and the derivative is zero throughout it. The mean value theorem therefore gives $f(q)=f(p)$. Since the two points were arbitrary, the function is constant on the interval.

For continuity at any $x$, choose $\delta ={\eta }^{1/\alpha }$ for a desired error $\eta >0$. If $|y-x|<\delta$, the stated bound gives $|f(y)-f(x)|<\eta$. Thus $f$ is continuous, in fact uniformly continuous.

When $\alpha >1$, the difference quotient satisfiesTherefore the derivative exists at every point and is zero. Part (i) proves $f$ is constant. This is the lemma that Hölder exponent greater than one forces constancy; no differentiability was assumed in advance.

Let $D=f_{+}^{\prime }(a)$ and write $g(t)=(f(t)-f(a))/(t-a)$ for $t>a$. By the assumed right derivative, choose $\delta >0$ with $|g(t)-D|<ϵ/2$ whenever $a<t<a+\delta$. Choose $b_0\in (a,\min (b,a+\delta ))$.

Apply the mean value theorem to $f$ on $[a,b_0]$. It gives $x\in (a,b_0)$ with $f^{\prime }(x)=g(b_0)$. Both $g(x)$ and $g(b_0)$ are within $ϵ/2$ of $D$, soThis proves the endpoint secant-tangent approximation without any continuity assumption on $f^{\prime }$. The continuity required by the theorem is only that of $f$ itself.