📂Fourier Analysis

The sufficient condition for the Fourier series of a function to converge absolutely and uniformly to the function

Theorem

The function $f$, defined in $[L, -L)$, is continuous and piecewise smooth. Therefore, the Fourier series of $f$ absolutely and uniformly converges to $f$.

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When $f$ is piecewise smooth, its Fourier series converges pointwise to $f$. If the condition that $f$ is continuous is strengthened by removing the discontinuity points of $f$, then the Fourier series of $f$ absolutely and uniformly converges to $f$. The proof uses the Cauchy-Schwarz inequality and Weierstrass M-test.

Weierstrass M-test

For functions $f_{n}$ and $z \in A$, if there exists a sequence of positive numbers $M_{n}$ satisfying $|f_{n}(z)| \le M_{n}$, and if $\displaystyle \sum_{n=1}^{\infty} M_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty} f_{n}(z)$ absolutely converges, and it converges uniformly in $A$.

Proof

The Fourier series of $f$ is $\sum c_{n}e^{i\frac{n\pi t}{L}}$, so it will be shown that this is greater than or equal to some $a_{n}$ for which $a_{n} < \infty$ holds.

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Based on the relationship between the Fourier coefficients of $f$ and the Fourier coefficients of its derivative

$$c_{n}=\frac{L}{in\pi}c_{n^{\prime}} \quad n\ne 0 \\ \implies |c_{n}|=|\frac{L}{n\pi} c_{n^{\prime}}|$$

And by the Bessel’s inequality

$$\begin{equation} \sum \limits_{n= -\infty}^{\infty} |c_{n^{\prime}}| \le \dfrac{1}{2L}\int_{-L}^{L}|f^{\prime}(t)|^2dt < \infty \label{eq1} \end{equation}$$

Now, by expressing $\sum |c_{n}|$ as $c_{n^{\prime}}$

$$\begin{align*} \sum \limits_{n= -\infty}^{\infty} |c_{n}| &= |c_{0}| + \sum _{n \ne 0} |c_{n}| \\ &= |c_{0}| + \sum _{n \ne 0} \left| \dfrac{L}{n\pi}c_{n^{\prime}} \right| \\ & \le & |c_{0}| + \left( \sum \limits_{n \ne 0} \dfrac{L^2}{n^2 \pi ^2} \right)^\frac{1}{2} \left( \sum \limits_{n \ne 0} |c_{n^{\prime}}|^2 \right)^{\frac{1}{2}} \\ &=|c_{0}| + \left( \dfrac{L^2}{\pi ^2}\sum \limits_{n \ne 0} \dfrac{1}{n^2 } \right)^\frac{1}{2} \left( \sum \limits_{n \ne 0} |c_{n^{\prime}}|^2 \right)^{\frac{1}{2}} \end{align*}$$

The second line holds by the Cauchy-Schwarz inequality, and using $\sum _{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6} <\infty$, by the formula $\eqref{eq1}$, the last line is

$$|c_{0}| + \left( \dfrac{L^2}{\pi ^2}\sum \limits_{n \ne 0} \dfrac{1}{n^2 } \right)^\frac{1}{2} \left( \sum \limits_{n \ne 0} |c_{n^{\prime}}|^2 \right)^{\frac{1}{2}} < \infty$$

Therefore, by the Weierstrass M-test, $\sum \limits_{-\infty}^{\infty}c_{n}$ absolutely and uniformly converges to $f$.

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