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let $f\in C[-\pi,\pi]$ i.e $f$ is continuous over $[-\pi,\pi]$

Evaluate:lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos(nt)\,\mathrm dt$

I tried to evaluate it using normal integration technique but that yielded nothing.Any hints

Another one is lim $_{h\rightarrow 0}\frac{1}{h}\int _{-h}^hf(t)\,dt $ How to do this?

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    $\begingroup$ The first one can be found using what is called the Riemann-Lebesgue lemma. I would recommend looking that up. For the second one, write $g(x)=\int_0^xf(t)\ dt$ and use the Fundamental theorem of calculus. $\endgroup$ – Jason Nov 26 '14 at 5:41
  • $\begingroup$ For 1, first try to calculate $\lim_n \int_{-\pi}^\pi \cos(nt) dt$. Then use the fact that $f$ is bounded since it is continuous on a compact set. $\endgroup$ – Xiao Nov 26 '14 at 5:55
  • $\begingroup$ @Xiao: actually, I think the argument is that continuous and compact imply uniform continuity, not boundedness (if that's a word). If the integral was over $(0,\pi]$ and $f(x)=\cos(1/x)$, you'd still have a continuous and bounded function, but it probably wouldn't work $\endgroup$ – Alexandre Halm Nov 26 '14 at 7:40
  • $\begingroup$ @Alex: It would work. $\endgroup$ – Jonas Meyer Nov 26 '14 at 8:40
  • $\begingroup$ @JonasMeyer: looks like it would indeed. What would be a good counter-example? If $f$ has bounded variation, we should have $|\hat f(n)| = O(\frac{1}{n})$, so any counter-example should not be BV ($\cos(1/x)$ isn't). $\endgroup$ – Alexandre Halm Nov 26 '14 at 9:33
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Trying to bring back old memories of my teacher's very good elementary proof ... I'll try to give a sketch and hopefully some intuition for 1. (2. is trivial as per @Jason's comment):

First note that your integral can be written as $$\int _{-\pi}^{\pi} f(t)\cos(nt)dt = \frac{1}{n} \sum_{k=-n}^{n-1} \int_{k\pi}^{(k+1)\pi} f(u/n)\cos(u)du = \\ = \frac{1}{n} \sum_k (-1)^k \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \tag{1}$$

Now when $n \rightarrow \infty$, $v \mapsto f\left(\frac{v+k\pi}{n}\right)$ becomes very "flat" over $[0,\pi]$ while over the same interval $\int_0^{\pi} \cos(v)dv = 0$. So intuitively you would like to see something like $$\int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \sim \int_0^\pi f\left(\frac{k\pi}{n}\right)\cos(v)dv =0$$

To get to that, take $\epsilon > 0$. Since $f$ is continuous on $[0,\pi]$, it is also uniformly continuous (from the Heine-Cantor theorem), so there is a $\eta = \eta_{\epsilon} > 0$ so that $\lvert x-y \rvert < \eta \implies \lvert f(x)-f(y) \rvert < \epsilon$.

Now let's take an integer $n > N_{\epsilon} = \tfrac{\pi}{\epsilon}$. You would have for all $k$'s: $$\forall v \in [0,\pi], \left| f\left(\frac{v+k\pi}{n}\right) - f\left(\frac{k\pi}{n}\right)\right| < \epsilon$$ and thus (oh what a mouthful of LaTeX ...):

$$ \left| \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \right| = \left| \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv - \int_0^\pi f\left(\frac{k\pi}{n}\right)\cos(v)dv \right| \le \\ \int_0^\pi \left| f\left(\frac{v+k\pi}{n}\right) - f\left(\frac{k\pi}{n}\right)\right| |\cos(v)|dv \le \epsilon \pi $$

Now inject that in $(1)$ and you get: $$ n\ge N_{\epsilon} \implies \left|\int _{-\pi}^{\pi} f(t)\cos(nt)dt \right| \le 2\pi\epsilon$$

And you conclude $$\lim_{n\rightarrow\infty} \int _{-\pi}^{\pi} f(t)\cos(nt)dt = 0$$

Now I hope I haven't ridiculed myself with some major mistake...

The good thing with this proof is that you actually see what topological and functional properties are required to get the result:

  • the interval $[0,\pi]$ is compact in $\mathbb R$
  • $f$ is continuous and thus uniformly continuous
  • $\cos$ verifies $\int \phi = 0$ and $\int |\phi| < \pi$ on each interval $[k\pi,(k+1)\pi]$

It's probably easy to find examples that show that relaxing any of these conditions would void the result.

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  • $\begingroup$ Thanks for including the intuition $\endgroup$ – Simon S Nov 26 '14 at 13:13
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Nice one Alex H. :) , I have a proof (that is, in terms of topological tools, close to yours) using a different implementation.

I use two results:

-Firstly, if the hypothesis on the function is $f\in C^1[-\pi,\pi]$ then you get the result with an integration by part on the cos, and deriving the function. Very easy to prove.

-Secondly, I'll use the Weirstrass theorem: there is a sequence of polynomials that uniformly converges $\rightarrow f$. I'll note the sequence $P_k$.

Now: $| \int_{-\pi}^{\pi} f(t)\cos(nt)dt - \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt | = |\int_{-\pi}^{\pi} (f(t)-P_k(t))\cos(nt)dt| \leq 2\pi |f-P_k|_{\infty}$

So, for a certain N positive integer, we get:

$ k \geq N: | \int_{-\pi}^{\pi} f(t)\cos(nt)dt - \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt | \leq \epsilon $ ; thanks to the uniform convergence of the ($P_k$)

From the first point we know that : $ \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt \rightarrow 0 $ , when $n \rightarrow +\infty$ , since every polynomial is $C^1[-\pi,\pi]$.

Using | |a| - |b| | $\leq |a-b|$ we get : $|\int_{-\pi}^{\pi} f(t)\cos(nt)dt| \leq \epsilon + |\int_{-\pi}^{\pi} P_k(t)\cos(nt)dt| \leq 2\epsilon $ , for k>N, n > $N_1$ set with the above result.

Hence you get what you wanted

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  • $\begingroup$ Actually that was my classroom proof at the beginning of the Fourier series chapter, in the last century. I'm wondering, my proof relies mostly on the Heine-Cantor theorem (so a property of compact sets in generic metric spaces - I think), while yours relies on Stone-Weierstrass, which is a property of Hausdorff spaces. Any idea where the connection is? I forgot most of these real topology constructions $\endgroup$ – Alexandre Halm Nov 26 '14 at 15:11
  • $\begingroup$ @Alex H. Yes, you prove the weierstrass theorem using the Heine-Cantor theorem :), that's the essential part in the proof. That's why i said both proofs were close, mine being the usual one I think $\endgroup$ – mvggz Nov 26 '14 at 15:14
  • $\begingroup$ Of course! Thanks for the refresh. $\endgroup$ – Alexandre Halm Nov 26 '14 at 15:30

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