Given a continuous function $f_0: [0,1] \rightarrow \mathbb{R}$, define
$$f_n(x) = \int^x_0 f_{n-1}(t) dt, x \in [0,1]$$
for $n=1,2,3,...$ .
For each $x \in [0,1]$, show that $\sum^{\infty}_{n=1} f'_n(x)$ converges.
Also, show that the function $$g(x) = \sum^{\infty}_{n=1} f'_n(x)$$ is continuous.

I'd like to prove this problem but I didn't answer even the first question. How to I know that $\sum^{\infty}_{n=1} f'_n(x)$ converges. Thanks in advance.

  • $\begingroup$ I assume your sum should start from $1$, not $0$ ($f_0$ is not given to be differentiable). $\endgroup$ – Ian Aug 5 '15 at 12:07
  • $\begingroup$ @lan You are right. $\endgroup$ – Jeong Aug 5 '15 at 12:09
  • $\begingroup$ Recall that if a sequence of continuous functions converges uniformly, then the limit function is continuous. $\endgroup$ – Math1000 Aug 5 '15 at 12:17

You have that

$$f'_n(x) = f_{n-1}(x)$$

Now let's prove by recurence that

$$|f_n(x)| \leq M \frac{x^n}{n!}$$

Indeed, $|f_0(x)| \leq M$ as it is continuous on $[0,1]$

Now if $|f_n(x)| \leq M \frac{x^n}{n!}$, then

$$|f_{n+1}(x)| = \left| \int_0^x f_n(t)dt \right| \leq \int_0^x \left| f_n(t)\right| dt $$

$$\leq M \int_0^x \frac{t^n}{n!} dt = M \frac{x^{n+1}}{(n+1)!}$$

And the property is proved

So you have

$$\sum_{n=1}^\infty \| f'_n\|_\infty \leq M\sum_{n=1}^\infty \frac{1}{n!} = Me$$

And your serie converge normaly. This gives you the pointwise convergence and the continuity of the serie


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.