Let be $f_n $ be a sequence of functions and $f_0$ an continuous arbitrary function derivable in $0$ such that: $$f_{n+1}\left(x\right)=\frac{1}{x}\int _0^xf_n\left(t\right)dt$$ for every $n$ positive integer.

The domain of $f_n$ is $[0,1]$

I was wondering if the following statement is true or not:

$f'_n $is uniformly convergent to the function:

$g(x)=0$ for every $x \in [0,1]$

I was thinking that I need to prove that:

$$sup_{x\in [0,1]}|f'_n(x)-f(x)|=0$$

that is to say for every $\epsilon>0$ there is a positive integer $N$ such that for every $n>N$ and ${x\in [0,1]}$ we have:

$$|f'_n(x)-f(x)|<\epsilon$$ How should I proceed?

  • $\begingroup$ What do you mean by a "random function"? $\endgroup$ May 14 at 11:38
  • $\begingroup$ an arbitrary function I will edit the question $\endgroup$
    – shangq_tou
    May 14 at 11:38
  • $\begingroup$ If $f_0\equiv 1$ then $f_n\equiv 1$ for all $n$. So the limit need not be $0$. $\endgroup$ May 14 at 11:39
  • $\begingroup$ I edited the question. I wanted to know of $f'_n$ is convergent to $0$. $\endgroup$
    – shangq_tou
    May 14 at 11:40

2 Answers 2


The operator $T(f) =1/x\int_0^x f$ is a continuous contraction on the space $\mathcal{C}_0[0,1]=\{f \in \mathcal{C}[0,1]: f(0) = 0\}$.The easy estimate $\lVert T(f) \rVert \leq \lVert f \rVert$ shows T is continuous. Evaluating $T$ on a polynomial $f(x) = a_1x\dots +a_n x^n$ yields the estimate $\lVert T(f) \rVert \leq 1/2\lVert f \rVert$ for polynomials. We apply Stone-Weierstrass to get the estimate for all $f$, which proves $T$ is a contraction with fixed point $f\equiv 0$

A few details: $\lVert \rVert$ is sup norm.

Define $T(f)(0) = 0$ (if this wasn't clear). Then L'Hôpital's rule give $T(f) $ continuous on $[0,1]$.

Also, S-W insures $\exists \, p_n\rightarrow f$, $p_n$ polynomials in $\mathcal{C}[0,1]$. Then if $f \in \mathcal{C}_0[0,1]$, $(p_n - p_n(0)) \rightarrow (f - f(0)) = f$, in sup norm, so we can assume $p_n \in \mathcal{C}_0$ to apply the estimate.


We can replace $f_n(x)$ by $f_n(x)-f_0(0)$ to reduce the proof to the case when $f_0(0)=0$. In this case, differentiabilty of $f_0$ at $0$ implies that there exists a constant $M$ with $|f_0(x) |\leq Mx$ for all $x$.

Now $xf_{n+1}'(x)+f_{n+1}(x)=(xf_{n+1}(x))'=f_n(x)$ so $|f_{n+1}'(x)|\leq \frac 1 x (|f_n(x)+|f_{n+1} (x)|)$. Use induction to show that $|f_n(x)| \leq \frac {Mx} {2^{n}}$ for all $n$. Hence, $f_{n+1}'(x) \to 0$ uniformly.

  • $\begingroup$ The base case that $|f_0(x)| \le Mx$ doesn't hold in general. Consider $f_0(x) = e^x$. $\endgroup$ May 14 at 12:13
  • $\begingroup$ @infinitylord Thanks for the comment. I have corrected the proof. $\endgroup$ May 14 at 12:20
  • $\begingroup$ The base case still fails, as if $M = \sup_{0 \le x \le 1}|f_0(x)|$ occurs at some $0 < x^\ast <1$, it follows that $|f_0(x^\ast)| = M > Mx^\ast$. However your conclusion holds if you take $M = \sup_{0\le x \le 1} g(x)$, where $g(x) = |f_0(x)|/x$ for $x \ne 0$ and $g(0) = f_0'(0)$. $\endgroup$ May 14 at 13:56
  • $\begingroup$ @infinitylord I just forgot to delete 'if $M=\sup_{0\leq x \leq 1} |f_0(x)|$ then '. I have already chosen $M$ in the beginning so this definition of $M$ should have been deleted. Yes, my $M$ is exactly what you mention in your comment, the supremum of $g$. $\endgroup$ May 14 at 23:14

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