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I need some help on the following problem.

Let $f\in L_1([-\pi,\pi])$. Then $\int_{-\pi}^\pi ~f(x) \cos (nx) \mathrm{d}\mu(x) \to 0$, where $\mu$ is the Lebesgue measure on $[\pi,\pi]$.

Any hints and suggestions on how to begin is very much welcomed. Thanks.

Bass
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    This is a special case of the Riemann-Lebesgue lemma: http://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma. Replace $f$ by $f 1_{[-\pi,\pi]}$ and take the real part. – Nate Eldredge Nov 27 '11 at 02:51
  • @Nate: How's the above a special case? – Bass Nov 27 '11 at 04:30
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    The Riemann-Lebesgue lemma says that for any integrable $g$, if we let $\hat{g}(t) = \int_\mathbb{R} g(x) e^{itx} dx$, then $\lim_{t \to \pm \infty} \hat{g}(t) = 0$. Take $g = f 1_{[-\pi,\pi]}$. Then $\mathop{Re} \hat{g}(t) = \int_{-\pi}^\pi f(x) \cos(tx)dx$, and your result follows. – Nate Eldredge Nov 29 '11 at 00:02

2 Answers2

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First, let's examine the functions $C_n(x)=\cos (nx)$. For large $n$, the graph of this function consists of many cosine waves of small common period. If you integrate this function over $[-\pi,\pi]$, you'll, of course, obtain 0.

Now if $I$ is any interval in $[-\pi,\pi]$, for $n$ big, the graph of $C_n$ over $I$ will consist of many cosine waves of small common period together with a portion of a cosine wave near each endpoint of $I$. Here, $\int_I C_n$ will be the same as integrating $C_n$ just "near the endpoints" of $I$ (the middle portion will integrate to 0). But as $n$ grows large, the measure of those portions gets small, and as a result, $\lim\limits_{n\rightarrow\infty} \int_I C_n =0$.

Thus, for any interval $I$ and for any number $a$, we have $$\lim_{n\rightarrow\infty} \int_I a\cdot C_n =0.$$

Using the above result and linearity of integration, we can show that if $g$ is a function of the form $$g(x)=\sum_{i=1}^n a_i \chi_{I_i},{\text{ where } }\ I_j\cap I_k\ \buildrel{j\ne k}\over=\ \emptyset{\text{ and }}\bigcup_{i=1}^n I_i=[-\pi,\pi],$$ then $$\lim_{n\rightarrow\infty} \int_{[-\pi,\pi]} g C_n =0.$$

Thus, the theorem is true for any step function in $L_1$.

The general result follows from the fact that the step functions are dense in $L_1$ (that is, given $f\in L_1$ and $\epsilon>0$, there is a step function $g$ with $\Vert f-g\Vert_{L_1}<\epsilon$).

I can provide more details here if you like, just let me know.

David Mitra
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  • Dear David, I was wondering if you had time to provide more details. I'm very much interested. Thanks. – Kuku Mar 26 '12 at 13:15
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This statement is true for any $f\in C^1([-\pi,\pi])$. Note that $C^1([-\pi,\pi])$ is dense in $L_1([-\pi,\pi])$ hence equality holds for all $f\in L_1([-\pi,\pi])$

Norbert
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