Recall that the Fourier coefficient is

$$ \begin{align*} \widehat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}\ dx \end{align*} $$

Here are a few more examples of finding the Fourier coefficients: Example 1, Example 2, Example 3 and Example 4. Next, recall that we defined the partial sum

$$ \begin{align*} S_Nf(x)=\sum_{n=-N}^{N}\widehat f(n)e^{inx} =\sum_{|n|\le N}\widehat f(n)e^{inx}. \end{align*} $$
Theorem (Absolutely Summable)
Suppose that \(\sum_{n=-\infty}^{\infty}|\widehat f(n)|<\infty\). Then \(S_Nf(x)\) converges uniformly to a continuous function.

So if the sum of the absolute value of the coefficients converges (in other words, absolutely summable), then the Fourier series converges uniformly to a continuous function. Though we don’t know if this continuous function is actually \(f\).

Proof

First note that since we don’t have a proposed limit, then we need to use the Uniform Cauchy criterion. Hence for all \(\epsilon > 0\), we want to show that there exists an \(N_0\) such that if \(M,N \ge N_0\), then

$$ \begin{align*} |S_Mf(x)-S_Nf(x)|<\varepsilon \end{align*} $$

for every \(x\). Next, take two partial sums \(S_M\) and \(S_N\) and assume without loss of generality that \(M>N\). Then expand the two partial sums:

$$ \begin{align*} S_Mf(x)=\sum_{|n|\le M}\widehat f(n)e^{inx}, \qquad S_Nf(x)=\sum_{|n|\le N}\widehat f(n)e^{inx}. \end{align*} $$

The terms with \(|n|\le N\) cancel, so

$$ \begin{align*} S_Mf(x)-S_Nf(x) = \sum_{N<|n|\le M}\widehat f(n)e^{inx}. \end{align*} $$

Now apply the triangle inequality:

$$ \begin{align*} |S_Mf(x)-S_Nf(x)| &= \left| \sum_{N<|n|\le M}\widehat f(n)e^{inx} \right|\\ &\le \sum_{N<|n|\le M} |\widehat f(n)e^{inx}| \quad \text{(by the triangle inequality)}\\ &= \sum_{N<|n|\le M} |\widehat f(n)|. \quad \text{(since $|e^{inx}| = 1$)} \end{align*} $$

By assumption, we know that \(\sum_{n=-\infty}^{\infty}|\widehat f(n)|<\infty\). Hence, for our chosen \(\varepsilon>0\), there exists \(N_0\) such that

$$ \begin{align*} \sum_{|n|>N_0}|\widehat f(n)|<\varepsilon. \end{align*} $$

Now take \(M>N\ge N_0\). Since the indices satisfying \(N<|n|\le M\) are part of the tail \(|n|>N_0\),

$$ \begin{align*} |S_Mf(x)-S_Nf(x)| &\le \sum_{N<|n|\le M}|\widehat f(n)|\\ &\le \sum_{|n|>N_0}|\widehat f(n)|\\ &<\varepsilon. \end{align*} $$

This holds for every \(x\), because the right-hand side does not depend on \(x\). Therefore, \(\{S_Nf\}\) is uniformly Cauchy. By the uniform Cauchy criterion, there exists a function \(g\) such that \(S_Nf\longrightarrow g\) uniformly. Next, note that each partial sum

$$ \begin{align*} S_Nf(x)=\sum_{|n|\le N}\widehat f(n)e^{inx} \end{align*} $$

is continuous because each \(e^{inx}\) is continuous, and a finite sum of continuous functions is continuous. Since \(S_Nf\to g\) uniformly and every \(S_Nf\) is continuous, then by the uniform limit theorem, \(g\) is continuous. \(\ \blacksquare\)

Fourier Coefficients Determine \(f\)

One question we can ask here is how do we know that \(g = f\)? Another related question is

Does \(\{ \widehat f(x)\} \) uniquely determine \(f\)?

To answer this, we have the following theorem:

Uniqueness Theorem of Fourier Series
Let \(f\) be an integrable function on \([0,2\pi]\) such that \(\widehat f(n) = 0\) for all \(n \in \mathbb{Z}\). Then \(f(x) = 0\) whenever \(x\) is a continuity point of \(f\).

Proof

TODO…

From this theorem, we want to conclude that if if \(\widehat{f}(n) = \widehat{g}(n)\) for all \(n \in \mathbb{Z}\). Then \(f\) and \(g\) are the same function. How do we apply it? Let \(h = f-g\). Observe that

$$ \begin{align*} \widehat{h}(n) = \widehat{f - g}(n) = \frac{1}{2\pi} \int_0^{2\pi} (f(x)-g(x)) e^{-inx} dx = \widehat{f}(n) - \widehat{g}(n) = 0 \end{align*} $$

Since \(\widehat h(n) = 0\) for all \(n \in \mathbb{Z}\), then by the theorem, we know that \(h(x) = 0\) at every continuity point of \(h\). Since \(h = f-g\), then \(f(x) - g(x) = 0\). Therefore

$$ \begin{align*} f(x) = g(x) \end{align*} $$

at every continuity point of \(f-g\). Note that if \(f\) and \(g\) are continuous, then \(f-g\) is continuous everywhere and so \(f = g\) everywhere. Next, we have

Theorem
Suppose that \(\sum_{n=-\infty}^{\infty}|\widehat f(n)|<\infty\) and \(f\) is continuous on \([0,2\pi]\). Then $$ \begin{align*} f(x) = \sum_{n \in \mathbb{Z}} \widehat f(n) e^{inx} \end{align*} $$

Some Notes: If \(\sum_{k\in\mathbb Z}\widehat f(k)e^{ikx}\) converges uniformly to a function \(g\), then \(g\) is continuous. Moreover, uniform convergence lets us interchange the sum and the integral which then gives us \(\widehat g(n)=\widehat f(n)\). Then by Uniqueness of the coefficients and the fact that \(g\) and \(f\) are continuous, we get that \(g = f\) everywhere. Recall that what gave us uniform convergence is actually Absolute summability.

Proof

Define

$$ \begin{align*} g(x) := \sum_{n\in\mathbb{Z}} \widehat{f}(n)e^{inx}. \end{align*} $$

We want to show \(g=f\). To do that, we compute the Fourier coefficients \(\widehat g(n)\) and we show that \(\widehat g(n)=\widehat f(n)\) for every \(n\in\mathbb Z\).

We know by assumption \(\sum_{n=-\infty}^{\infty}|\widehat f(n)|<\infty\). Hence by the absolutely summable theorem, \(g\) converges uniformly on \([0,2\pi]\). In particular, \(g\) is continuous. For each \(n\in\mathbb{Z}\), we want to compute

$$ \begin{align*} \widehat{g}(n) &= \frac{1}{2\pi}\int_0^{2\pi}g(x)e^{-inx}\,dx \\ &= \frac{1}{2\pi}\int_0^{2\pi} \left(\sum_{k\in\mathbb{Z}}\widehat{f}(k)e^{ikx}\right) e^{-inx}\,dx \\ \end{align*} $$

First we move \(e^{inx}\) into the sum since it doesn’t depend on the sum’s index.

$$ \begin{align*} \widehat{g}(n) &= \frac{1}{2\pi}\int_0^{2\pi} \sum_{k\in\mathbb Z} \widehat f(k)e^{ikx}e^{-inx}\,dx \\ \end{align*} $$

Because the series converges uniformly, we may interchange the integral and the infinite sum.

$$ \begin{align*} \widehat{g}(n) &= \frac{1}{2\pi} \sum_{k\in\mathbb Z}\widehat f(k) \int_0^{2\pi}e^{ikx}e^{-inx}\,dx. \\ &= \frac{1}{2\pi} \sum_{k\in\mathbb{Z}}\widehat{f}(k) \int_0^{2\pi}e^{i(k-n)x}\,dx \end{align*} $$

Recall that

$$ \begin{align*} \frac{1}{2\pi}\int_0^{2\pi}e^{i(k-n)x}\,dx = \begin{cases} 1, & k=n,\\ 0, & k\neq n. \end{cases} \end{align*} $$

Hence, every term is zero except the term where \(k=n\). Therefore,

$$ \begin{align*} \widehat g(n)=\widehat f(n). \end{align*} $$

Thus \(f\) and \(g\) have the same Fourier coefficients. Furthermore, by the uniqueness theorem for Fourier series, \(f=g\) almost everywhere. Since both \(f\) and \(g\) are continuous, they must agree everywhere. Hence

$$ \begin{align*} f(x)=\sum_{n\in\mathbb{Z}}\widehat{f}(n)e^{inx} \qquad\text{for every }x\in[0,2\pi]. \end{align*} $$

Piecewise Continuous?

Note that if the function is piecewise continuous so it’s continuous except possibly at finitely many points. Then we can still define

$$ \begin{align*} g(x)=\sum_{n\in\mathbb Z}\widehat f(n)e^{inx} \end{align*} $$

and assuming the coefficients are absolutely summable, we can still show that \(\widehat g(n)=\widehat f(n)\). However, the uniqueness theorem now gives \(f(x)=g(x)\) almost everywhere instead of everywhere.

More Question?

Recall that we first defined

$$ \begin{align*} S_N(x)=\sum_{n=-N}^{N}\widehat f(n)e^{inx}. \end{align*} $$

Absolute summability gives

$$ \begin{align*} S_N\longrightarrow g \quad\text{uniformly} \end{align*} $$

for some continuous function \(g\). We then proved that \(g=f\) by Uniqueness and the fact that \(f\) is continuous. Only after that, we concluded that \(S_N\longrightarrow f\). Now we have two questions

Question 1: When are the Fourier coefficients absolutely summable?

A common sufficient condition is that \(f\) is sufficiently smooth. For example, a \(2\pi\)-periodic continuously differentiable function has absolutely summable Fourier coefficients. We have to show this …

Question 2: Does continuity of \(f\) imply absolute summability?

The answer is no.

References

  • SFSU Fourier Analysis by Professor Chun-Kit Lai
  • Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi