Converse of mean value theorem

I am wondering if the following converse (or modification) of the mean value theorem holds. Suppose $f(\cdot)$ is continuously differentiable on $[a,b]$. Then for all $c \in (a,b)$ there exists $x$ and $y$ such that $$f'(c)=\frac{f(y)-f(x)}{y-x}$$

Assume that $f''(c)\ne0$. Then one can find $x_1<c<x_2$ such that $$f'(c)={f(x_2)-f(x_1)\over x_2-x_1}\ .$$ Proof. Assume $f''(c)>0$ and consider the auxiliary function $$g(t):=f(c+t)-f(c)-t f'(c)\ ,\tag{1}$$ which is defined in a full neighborhood of $t=0$. By Taylor's theorem one has $$g(t)=g(0)+t g'(0)+{t^2\over2}g''(0)+o(t^2)={t^2\over2}\bigl(f''(c)+o(1)\bigr)\qquad(t\to0)\ .$$ It follows that there is an $h>0$ with $g(t)>0$ for $0<|t|\leq h$. Assuming $g(h)\geq g(-h)$ put $t_1:=-h$, and choose $t_2\in\ ]0,h]$ such that $g(t_2)=g(t_1)$, which is possible by the intermediate value theorem.

Finally put $x_i:=c+t_i$ $(i=1,\>2)$. Then it follows from $(1)$ that $$f(x_2)-f(x_1)=g(t_2)-g(t_1)+(t_2-t_1)f'(c)=(x_2-x_1) f'(c)\ ,$$ which is equivalent to the claim.

• Do we need the restriction that $g(-h) \neq g(h)$? May 1, 2014 at 10:08
• @user103828: The function $g$ is not symmetric with respect to $t=0$; therefore we cannot expect $g(-h)=g(h)$. This is not a restriction but a fact we have to reckon with somehow. See my edit. May 1, 2014 at 10:50

It is not true. Let $f(t)=t^3$. Then $f'(0)=0$, but $\frac{f(y)-f(x)}{y-x}$ is never $0$.

• Thanks. It seems also from your example that there is no easy'' way to modify the statement to make it true. May 1, 2014 at 6:35
• You are welcome. There is always a way, but offhand I cannot think of one. May 1, 2014 at 6:37
• What if it is modified such that the derivative is nowhere vanishing? Jan 27, 2016 at 22:54
• I am answering quickly, busy with other stuff, so may be wrong. But it looks to me as if $t^3+17$ will also give a counterexample. Jan 27, 2016 at 23:01
• $t^3 + t$ is a counterexample. @MathematicsStudent1122 Feb 8, 2018 at 23:39