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.
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 $t\le 1$.)