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I practise before the real analysis exam and I got stuck on few excercises.

"if $F$ is an antiderivative of a continuous periodic function and $F$ is bounded, then $F$ is periodic"

$~\to~$ one should give a proof or counterexample.

(I'm convinced it's true, but how to prove it?)

Matcha Latte
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larry01
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    I don't know how much you've tried to prove this (you don't suggest any particular approach), but one way to get started would be to pick a simple periodic function and check. Hint: Pick the simplest nonzero periodic function (and I mean simplest). – hardmath Feb 10 '15 at 01:11
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    You may have some notion of the domain of the function that is clear to you, but would benefit the Readers if made explicit. – hardmath Feb 10 '15 at 01:13

2 Answers2

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Hint(s):

Start with the definition of the antiderivative(s) of $f$:

$$F(x) = \int_0^xf(t)dt + C$$

where $C$ is a constant. Since $f$ is periodic, with period $T>0$, then

$$F(2T) = \int_0^{2T}f(t)dt + C = \int_0^{T}f(t)dt + \int_T^{2T}f(t)dt + C = 2\int_0^{T}f(t)dt + C$$ $$F(3T) = \int_0^{3T}f(t)dt + C = F(2T) + \int_T^{2T}f(t)dt = 3\int_0^{T}f(t)dt + C$$

Can you generalize this to find an expression for $F(kT)$ for any integer $k$ (for example by induction)? What does this tell you about $\int_0^{T}f(t)dt$ given the conditions $F$ have to satisfy?

Finally what is the value of

$$F(x+T) - F(x) = \int_x^{x+T}f(t)dt = \int_x^{T}f(t)dt + \int_T^{x+T}f(t)dt$$

given the result found above?

Winther
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Hint:
What is special about $1 + \sin(x)$ that gives it the following property? $$\lim_{R\to\infty} \int_0^{R}1 + \sin(x) \ dx = \infty$$

If this integral was to have a maximum, what points would you look for first?

Further, $$\frac{d}{dR}\int_0^{R}f(x) dx = f(R)$$

jameselmore
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