A curious theorem by Peano Let $f$ be defined on $[a,b]$ and there differentiable.
Show that for every $ \epsilon>0 $ there exists a partition $\, a=a_0<a_1<...<a_n=b \,$ of $ \,[a,b] \,$ so that $$\left|\frac {f(a_{i+1})-f(a_i)}{a_{i+1}-a_i}-f'(a_i) \right|<\epsilon\qquad(i=0,...,n-1)$$
This proof was left as an exercise to the Belgian mathematician P.Gilbert by G.Peano in a quarrel (1884) about a mistake made by C.Jordan in his Cours d'analyse vol.1 (1882).
According to Peano, Jordan's proof of the mean value inequality theorem presented a fallacious argument: Gilbert did not agree. See nice pp. 12-14.
Can someone give her/his own proof ?
References


*

*T.M.Flett  $\,$Some Historical Notes and Speculations concerning the Mean Value Theorems of the Differential Calculus (1974) Bull.Inst.Math.Appl. vol.10 pp.66-72 (there is a proof by Flett)

*J.Mawhin $\,$Some Contributions of Peano to Analysis in the Light of the Work of Belgian Mathematicians in F.Skof (Ed.) Giuseppe Peano between Mathematics and Logic (2011) Springer (there is Gilbert's proof)

 A: Note that by choosing $x=a_{n-1}$, Peano's theorem has the following immediate consequence, for any $a < b$ and $\epsilon > 0$.
Lemma 1: If $f\colon[a,b]\to\mathbb{R}$ is differentiable, then there exists $x\in[a,b)$ satisfying
$$
\left\lvert\frac{f(b)-f(x)}{b-x}-f^\prime(x)\right\rvert\lt\epsilon.
$$
This lemma is obviously true for continuously differentiable functions simply by taking $x$ to be close enough to $b$. For functions with non-continuous derivatives, however, things are more difficult, and the inequality can fail for $x$ arbitrarily close to $b$. Starting with this lemma, we can also go in the opposite direction, and prove Peano's theorem, which I will do now. I'll also give a proof of the Lemma, showing that Peano's theorem is indeed true.
First, the natural way to try and prove Peano's theorem is to start with $a_0$ and then inductively choose values for $a_1,a_2,\ldots$. Once $a_i$ has been chosen then, if $a_i=b$ we are done. If not then, by the definition of the derivative of $f$ at $a_i$, there does exist $a_{i+1}\in(a_i,b]$ satisfying the required inequality. In order to maximise the chance of covering the interval $[a,b]$ in a finite number of steps, it would seem to be prudent to choose $a_{i+1}$ as large as possible and $a_{i+1}=b$ if possible. Unfortunately, this approach does not always work, and it is possible that you just end up generating an infinite increasing sequence $a_i$ in the interval $[a,b)$.
A method that does succeed is to, instead, work from the right to the left. That is, start by choosing $b_0=b$ and inductive define a decreasing sequence $b_0\gt b_1\gt b_2\gt\cdots\gt b_n=a$ such that
$$
\left\lvert\frac{f(b_i)-f(b_{i+1})}{b_i-b_{i+1}}-f^\prime(b_{i+1})\right\rvert\lt\epsilon.
$$
Then Peano's theorem follows by taking $a_i=b_{n-i}$.
If, for any $x\in(a,b]$ we let $S_x$ be the set of $y\in[a,x)$ with $\lvert(f(x)-f(y))/(x-y)-f^\prime(y)\rvert\lt\epsilon$ then, Lemma 1 applied to the interval $[a,x]$ says that $S_x$ is nonempty. Choosing a sequence $\delta_0,\delta_1,\ldots$ of positive reals tending to zero, select $b_0,b_1,\ldots$ as follows.


*

*$b_0=b$.

*Once $b_i$ has been chosen, if $b_i > a$ then take $b_{i+1}\in S_{b_i}$ with $b_{i+1}\le\inf S_{b_i}+\delta_i$. If possible, choose $b_{i+1}=a$.

*If $b_i=a$ then we are done.


Note that, as Lemma 1 implies that $S_{b_i}$ is non-empty whenever $b_i > a$, then this process continues either indefinitely or until $b_i=a$.
Lemma 2: The sequence $\{b_i\}$ terminates with $b_n=a$ for some $n$.
Proof: (Note: This proof will assume Lemma 1) Using proof by contradiction, suppose that the sequence $b_0,b_1,\ldots$ does not terminate. In this case, it decreases to a limit $c\in[a,b)$. If $c=a$ then, by the definition of the derivative at $a$, we have $\lvert (f(b_i)-f(a))/(b_i-a)-f^\prime(a)\rvert\lt\epsilon$ for large $i$. By the procedure above, we would choose $b_{i+1}=a$ and the sequence terminates, contradicting the assumption.
If $c > a$ then, using Lemma 1 applied to the interval $[a,c]$, there exists an $x\in[a,c)$ with $\lvert(f(c)-f(x))/(c-x)-f^\prime(x)\rvert\lt\epsilon$. By the continuity of $f$ at $c$, we have $\lvert(f(b_i)-f(x))/(b_i-x)-f^\prime(x)\rvert\lt\epsilon$ for all large $i$, so $x\in S_{b_i}$. By the procedure above, we choose $b_{i+1}\lt x + \delta_i$. Taking $i$ large enough that $\delta_i\le c-x$ this gives $b_{i+1}\lt c$, contradicting the property that $b_i$ decreases to $c$.
QED
To finish off the post:
Proof of Lemma 1:
By the definition of the derivative at $b$, there exists $\delta\gt0$ such that $\lvert(f(b)-f(x))/(b-x)-f^\prime(b)\rvert\lt\epsilon/2$ whenever $x\gt b-\delta$. Choosing $y\in[a,b)$ with $y\gt b-\delta$, applying the mean value theorem gives $x\in(y,b)$ such that $f^\prime(x)=(f(b)-f(y))/(b-y)$. Then,
$$
\begin{align}
\left\lvert\frac{f(b)-f(x)}{b-x}-f^\prime(x)\right\rvert
&=
\left\lvert\frac{f(b)-f(x)}{b-x}-\frac{f(b)-f(y)}{b-y}\right\rvert\cr
&\le
\left\lvert\frac{f(b)-f(x)}{b-x}-f^\prime(b)\right\rvert
+
\left\lvert\frac{f(b)-f(y)}{b-y}-f^\prime(b)\right\rvert\cr
&\lt\epsilon/2+\epsilon/2=\epsilon.
\end{align}
$$
QED
So, Peano's theorem is true. However, I made use of the mean value theorem in the proof of Lemma 1, so this does not give a method of proving the mean value inequality. That would be circular. The real problem is to prove Peano's theorem without making use of anything as strong as the mean value inequality. However, in my opinion, Peano's theorem is more difficult to prove than the MVT (or an approximate form of it), so does not give a useful step in proving the mean value inequality.
