5
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.