I am trying to prove that if $f\in\mathcal{C}^m([0,1])$, then the Fourier coefficients of $f$, $$\hat{f}(k)=\int_{0}^1f(x)e^{2\pi i k x}dx,$$ satisfy $$\hat{f}(k)=o\left(\frac{1}{k^m}\right).$$ I have tried to do it via integration by parts but I don't get anywhere.
Fourier Coefficients
1
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integration
functional-analysis
fourier-series
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1It looks like you want to integrate by parts $m$ times, passing the derivative onto $f^{(l)}$ and integrating the exponential every time. Have you tried this? – 2017-01-09
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0Do you have a periodic boundary condition? – 2017-01-09
1 Answers
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Integrating by parts with $\displaystyle u=f(x)$ and $\displaystyle v=\frac{e^{i2\pi kx}}{i2\pi k}$ yields
$$\int_0^1 f(x)e^{i2\pi kx}\,dx=\frac{f(1)-f(0)}{i2\pi k}-\frac{1}{i2\pi k}\int_0^1 f'(x)e^{i2\pi kx}\,dx $$
Repeated integration by parts with $\displaystyle u=f^{(n)}(x)$ and $\displaystyle v=\frac{e^{i2\pi kx}}{i2\pi k}$ reveals
$$\int_0^1 f(x)e^{i2\pi kx}\,dx=\sum_{n=0}^{m-1} (-1)^n\frac{f^{(n)}(1)-f^{(n)}(0)}{(i2\pi k)^{n+1}}+(-1)^m\frac{1}{(i2\pi k)^m}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx \tag 1$$
If $f$ is periodic, then all of the terms in the summation in $(1)$ are zero and we have
$$\int_0^1 f(x)e^{i2\pi kx}\,dx=(-1)^m\frac{1}{(i2\pi k)^m}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx$$
Since the Riemann-Lebesgue Lemma guarantees that $\lim_{k\to \infty}\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx=0$, then
$$\int_0^1 f^{(m)}(x)e^{i2\pi kx}\,dx=o(1)$$
as $k\to \infty$ and we have the desired result.