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Assume that $f:[0,\pi] \to \mathbb{R}$ is a continuously differentiable function.

I've been trying to prove that $$ \frac1n \int_0^\pi f'(x) \cos(nx)\, dx$$ tends to 0 and n tends to infinity.

To do this, I've been told that I must show the following:

$$ \left| \frac1n \int_0^\pi f'(x) \cos(nx)\, dx\right| \le \frac1n \int_0^1 \left|f'(x)\right| dx ≤ \frac1n \sup_{[0,1]}\left|f'(x)\right| \to 0 $$ as $n$ tends to $\infty$.

However, I'm struggling to see how to do this.

This is how far I've gotten:

$$ \left| \frac1n \int_0^\pi f'(x) \cos(nx) dx\right| \le \left| \frac1n \int_0^\pi f'(x)\, dx \right| \le \frac1n \int_0^\pi \left|f'(x)\right|\, dx $$

From this point onwards, I'm unsure to how you can say that $ \frac1n \int_0^\pi |f'(x)| dx \le \frac1n \int_0^1 |f'(x)| dx $. If you don't know what $f'(x)$ is explicitly, then how can you assume that the integral between 0 and 1 is larger?

I am also unsure as so why you can say that $ \frac1n \int_0^1 |f'(x)| dx ≤ \frac1n \sup_{[0,1]}|f'(x)| $, but this is more likely due to my lack of understanding of supremum.

Can someone explain why these inequalities hold?

Thank you.

M.Mass
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mimyo
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  • Since $f'$ is continuous, the quantity $L=\sup_{x \in [0,\pi]} |f'(x)|$ is finite hence ${\pi \over n} L \to 0$. What part are you having difficulty with? – copper.hat Nov 12 '17 at 19:39
  • In general, if $f(x) \le g(x)$ then $\int f(x) dx \le \int g(x)dx$. You have $|f'(x) | \le L$ so $\int |f'(x)| dx \le \int L dx = L \pi$. – copper.hat Nov 12 '17 at 19:48
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    What if it`s a typo, and it has to be $\sup_{[0, \pi]} |f^{´}(x)|$, would that be a problem? – Verbe Nov 12 '17 at 19:29
  • @Verbe I feel like this was the issue! It would make much more sense if it was π instead of 1. I'll confirm this with my lecturer. Thank you! – mimyo Nov 12 '17 at 21:28
  • @copper.hat I was confused as to why the inequality was valid when the limit changed from π → 1. As mentioned by Verbe, it would make sense if this was just a typo. Thank you for your clarification about the supremum though! – mimyo Nov 12 '17 at 21:29
  • It must be a mistake. You can choose some $f'$ such that the inequality does not hold. – copper.hat Nov 12 '17 at 21:59

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