Number of times a continuous function changes sign in an interval This is problem 7.23 in Apostol's Mathematical Analysis book.
Suppose $f$ is continuous on $[0,a]$. Let $f_0(x)=f(x)$ and
$$ f_{n+1}(x)=\frac{1}{n!}\int^x_0(x-t)^nf(t)\,dt,\qquad n=0,1,\ldots $$
(a) Show that $f^{(n)}_n(x)=f(x)$
(b) (Attributed to Fekete) Prove that the number of times $f$ changes sign in $[0,a]$ is at least the number of changes in sign if the ordered sequence $f(a),f_1(a),\ldots,f_n(a)$ for any $n\in\mathbb{N}$.
(c) (Attributed to Fejér) Use (b) to show that the number of times $f$ changes sign in $[0,a]$ is at least the number of changes of sign in the ordered sequence
$$f(0),\quad\int^a_0f(t)\,dt,\quad,\ldots,\quad \int^a_0t^nf(t)\,dt$$

I got part (a) by  applying  the chain rule to composition $F\circ\phi$ where $F(x,y)=\frac{1}{n!}\int^x_0(y-t)^n\,f(t)\,dt$, and $\phi(x)=(x,x)$. In fact, we get
$f'_{n+1}(x)=f_n(x)$, and $f'_1(x)=f(x)$.
It seems part(c) follows from part (b) by using $g(x)=f(a-x)$ in place of $f$.

Part(b) is giving me some problem. Any hint will be appreciated.
 A: Notice that

*

*the factor $t\mapsto (x-t)^n \mathbb{1}_{[0,x]}$,
$n\in\mathbb{Z}_+$, $0\leq x\leq a$, in the integrand is positive in $(0,x)$.


*$f_{n+1}(x)=\frac{1}{(n-1)!}\int^x_0(x-t)^{n-1} f_1(x)\,dx$
We now argue by induction.
Suppose $f(a)f_1(a)<0$. Since $f_1(a)=\int^a_0 f(t)\,dt$, $f$ must have at least one change of variable. This means that the statement in (b) holds for $n=1$,
Assume the statement in (b) holds for $n-1$ ($n\geq2$). Suppose  that the number of changes of sign in the tuple $(f_1(a),\ldots,f_n(a))$ is $k$. Then by assumption, $f_1$ changes sign at least $k$ times.  Then, there are $j\geq k$ points $0<x_1<\ldots<x_j<a$ and $\{\delta_\ell>0:\ell=1,\ldots,j\}$ such that

*

*$f_1(x_{\ell})=0$, $\ell=1,\ldots,j$,

*$[x_\ell-\delta_\ell,x_\ell +\delta_\ell]\cap [x_{\ell+1}-\delta_{\ell+1},x_{\ell+1} +\delta_{\ell+1}]=\emptyset$ for $\ell=1,\ldots,j-1$

*$\min_{[x_\ell-\delta_\ell,x_\ell+\delta_j]}f_1(x)<0<\max_{[x_\ell-\delta_\ell,x_\ell+\delta_j]}f_1(x)$, $\ell=1,\ldots,j$.

Claim I:
If $f(a)$ and $f_1(a)$ have the same sign, then,  as $f'_1=f$ and $f_1(0)=0$,  $f$ changes sign on each interval $(0,x_1),\,(x_1,x_2),\ldots,\, (x_{j-1},x_j)$ at least once. Thus $f$ changes sign at least $j$ times.
Claim II:
Suppose $f(a)f_1(a)<0$. have opposite signs.
The same argument in claim I shows that $f$ has at least one change of sign in each interval $(x_\ell,x_{\ell+1})$, $\ell =1,\ldots,j$.
We show that $f$ changes sign in $(x_j,a)$ at least once. It is enough to consider the case where the $f_1$ does not change sign in $(x_j,a)$. Without loos of generality, assume $f_1\geq0$ in $(x_j,a)$ so that $f_1(a)>0$. Then, $f(a)<0$ and by continuity, there  is exists  $\delta_{j+1}>0$ such that $x_j+\delta_j<a-\delta_{j+1}$ and $f<0$ in $[a-\delta_{j+1},a]$, which  means that $f_1$ is strictly decreasing in $[a-\delta_{j+1},a]$. Since $f_1(x)=\int^x_{a_j}f(t)\,dt$, it follows that $f$ must change signs at least once in $(x_j,a-\delta_{j+1})$. Therefore, $f$ changes sign at least $j+11$ times.
This completes the induction argument.

Notes:
The key part in the argument is the fact that if $f$ is continuous and $f\geq0$ in an interval $I=[\alpha,\beta]$,  and  $\max_{x\in I}f(x)>0$, then in any  $\int_If>0$ and $x\mapsto \int^x_\alpha$ is monotone nondecreasing in $I$.
