# Can we have a continuous choice in the mean value theorem

Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be differentiable. Must there exist a continuous function $g:\{(a,b)\in \mathbb{R}^2: a<b\}\rightarrow \mathbb{R}$ such that:

For every two distinct real numbers $a,b$ (with $a<b$) we have: $g(a,b)\in [a,b]$ and $f'(g(a,b))=\frac{f(b)-f(a)}{b-a}$

If the answer is no, does it at least hold if we add the condition that $f$ is continuously differentiable ?

The mean value theorem (along with axiom of choice) guarantees that such a function $g$ must exist if we don't insist on continuity of $g$.

Thank you.

• I'm inclined to believe this is a duplicate. Commented Mar 5, 2016 at 13:46
• If you want continuous choice, just use some variant of implicit function theorem. Commented Mar 5, 2016 at 13:59
• @user251257 Nope. You might be able to show that a continuous choice exists in some neighborhood of any given point that way, but a simple example shows you can't fit those local choices together into a global continuous choice. Commented Mar 5, 2016 at 14:13
• @DavidC.Ullrich there are global variant of implicit function theorem. I haven't intended to say that the given assumptions are sufficient. Commented Mar 5, 2016 at 14:14
• So your point was that a continuous $g$ exists in those cases in which a continuous $g$ exists? Commented Mar 5, 2016 at 14:16

No, not even if $f$ is continnuously differentiable.
Consider a continuously differentiable $f$ so $f\ge0$, $f(0)=0$, $f'>0$ on $(0,1)$, $f'<0$ on $(1,2)$, $f(2)=0$, $f'>0$ on $(2,3)$, and $f(3)/3$ is larger than the maximum of $f'$ on $[0,1]$. Suppose we have a continuous $g$.
Now $g(0,2)=1$. For small $\epsilon>0$, $f(2+\epsilon)/(2+\epsilon)>0$, so $f'(g(0,2+\epsilon))>0$, so $0<g(0,2+\epsilon)<1$, since $f'<0$ at points near $1$ but to the right of $1$. As $x$ increases from $2+\epsilon$ to $3$ we have $f(x)/x>0$, hence $g(0,x)\ne1$ since $f'(1)=0$. So $g(0,x)$ is trapped in the interval $[0,1]$ as $x$ increases from $2+\epsilon$ to $3$, and hence $f'(g(0,3)) < f(3)/3$, contradiction.
Edit In stating that $g(0,2)=1$ I was assuming that the condition on $g$ was $g(a,b)\in(a,b)$, as in MVT. Looking again, I see you're assuming only $g(a,b)\in[a,b]$. Doesn't matter - it's still easy to rule out $g(0,2)=2$ by arguing as above, and if $g(0,2)=0$ more or less the same argument works (maybe we want to assume that $f(3)/3$ is larger than the supremumm of $f'(t)$ for $t\le 1$.)
• We can't have $g(0,2)=2$ because if we look at point $x=2-\epsilon$, then due to MVT it holds $0<g(0,2-\epsilon)<2-\epsilon$. If $1<g(0,2-\epsilon)<2$ then we see that $\frac{f(2-\epsilon)}{2-\epsilon}=f'(g(0,2-\epsilon))<0$ which is a contradiction. If $<g(0,2-\epsilon)<1$ then $\frac{f(2-\epsilon)}{2-\epsilon}=f'(g(0,2-\epsilon))>0$. However, this shows that $g$ must necessarily jump from $g(0,2)=2$ to $<g(0,2-\epsilon)<1$. Is this correct? Commented Nov 17, 2021 at 15:42