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This is from Carothers 14.38

If $f \in BV[0,2\pi], f(0)= f(2\pi)$ show that $\int_0^{2\pi}f(x)\sin(nx)dx$ exist for each $n$ natural and $$\left|\int_0^{2\pi}f(x)\sin(nx)dx\right| \leq \frac{V_0^{2\pi}f}{n}$$

To show that the integral is bounded we can see that:

$$\int_0^{2\pi}f(x)\sin(nx)dx= f(2\pi)sin(2n\pi)-f(0)sen(0) + \frac{1}{n}\int_0^{2\pi}f'(x)\cos(nx)dx = \frac{1}{n}\int_0^{2\pi}f'(x)\cos(nx)dx$$

And then:

$$\left|\int_0^{2\pi}f(x)\sin(nx)dx\right|=\left|\frac{1}{n}\int_0^{2\pi}f'(x)\cos(nx)dx\right|\leq \frac{1}{n}\int_0^{2\pi}\left|f'(x)\cos(nx)\right|dx \leq \frac{1}{n}\int_0^{2\pi}\left|f'(x)\right|dx=\frac{V_0^{2\pi}f}{n}$$

But I don't know how to prove the existence of the integral in the first place. Could you guys help me?

Martin Sleziak
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Trash Panda
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1 Answers1

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Fix $n$. Then first, $f(x) \sin nx$ is a function of bounded variation, since both $f$ and $\sin (nx)$ are (see here). Second, functions of bounded variations are Riemann integrable, so the integral exists (see here).

So we restrict our attention to proving the inequality $$\left|\int_0^{2\pi}f(x)\sin(nx)dx\right| \leq \frac{V_0^{2\pi}f}{n}$$ you used integration by part, which might not be applicable since $f$ might not even by continuous. But we can use a very similar idea.

Let $\mathcal P : 0=a_0 < a_1 < a_2 < \cdots a_k = 2\pi$ be any partition of $[0, 2\pi]$. Then since $f(x) \sin nx$ is Riemann integrable, we can use any tagged partition with $\Delta \mathcal P \to 0$ to approximate the integral. By the Mean Value Theorem, there is $b_j \in (a_{j-1}, a_j)$ so that

$$ \frac{g(a_j)-g(a_{j-1})}{a_j - a_{j-1}} = \sin (nb_j), \text{ where } g(x) = -\frac 1n \cos (nx).$$

Thus

\begin{align*} \sum_{j=1}^k f(b_j) \sin (nb_j) (a_j - a_{j-1}) &= \sum_{j=1}^k f(b_j) (g(a_j) - g(a_{j-1}))\\ &= f(b_1)(g(a_1) - g(a_0)) + f(b_2)(g(a_2) - g(a_1)) + \cdots f(k) (g(a_k ) - g(a_{k-1})) \\ &= -f(b_1) g(a_0) + f(b_k) g(a_k) + \sum_{j=1}^{k-1} g(a_j) (f(b_j) - f(b_{j+1})) \\ &= \frac 1n (f(b_1) - f(b_k))+ \sum_{j=1}^{k-1} g(a_j) (f(b_j) - f(b_{j+1})) \\ &\le \frac 1n (f(b_1) - f(b_k)) + \frac 1n \sum_{j=1}^{k-1} |f(b_j) - f(b_{j+1})|\\ &= \frac 1n (f(b_1)- f(0) + (f(2\pi) - f(b_k))) + \frac 1n \sum_{j=1}^{k-1} |f(b_j) - f(b_{j+1})| \end{align*} here in the last step we used $f(0 ) = f(2\pi)$. So

\begin{align*} \sum_{j=1}^k & f(b_j) \sin (nb_j) (a_j - a_{j-1}) \\ &\le \frac 1n \left( |f(b_0)-f(0)| + |f(b_1)- f(b_0)| + \cdots +| f(b_k)- f(b_{k-1})| + |f(2\pi) - f(b_k)|\right)\\ & \le \frac{V^{2\pi}_0 (f) }{n}. \end{align*} (We are thinking of $0 < b_0 < \cdots< b_k < 2\pi$ as a partition of $[0, 2\pi]$). thus by taking any such tagged partition and let $\Delta \mathcal P \to 0$ (that is, take a sequence $\mathcal P_q$ of such tagged partition with $\Delta \mathcal P_q \to 0$ as $q\to \infty$), we obtain

$$ \int_0^{2\pi} f(x) \sin (nx) dx \le \frac{V^{2\pi}_0(f)}{n}.$$

Arctic Char
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