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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.

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  • $\begingroup$ 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
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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

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