# Let $f_n$ be a$f_{n+1}\left(x\right)=\frac{1}{x}\int _0^xf_n\left(t\right)dt\:$

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?

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

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.

• The base case that $|f_0(x)| \le Mx$ doesn't hold in general. Consider $f_0(x) = e^x$. May 14 at 12:13
• @infinitylord Thanks for the comment. I have corrected the proof. May 14 at 12:20
• 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 $g(0) = f_0'(0)$. May 14 at 13:56
• @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 $g$. May 14 at 23:14