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?

  • 2
    $\begingroup$ 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

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.

  • $\begingroup$ Thanks for including the intuition $\endgroup$ – Simon S Nov 26 '14 at 13:13

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

  • $\begingroup$ 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

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.