Convergence of Fourier Series

Most of the results presented for $2\pi$-periodic functions extend easily to functions 2L-periodic functions. So we only discuss the case of $2\pi$-periodic functions.

Definition. The function f(x) defined on [a,b], is said to be piecewise continuous if and only if, there exits a partition $\{x_1,x_2,..,x_n\}$ of [a,b] such that

(1)

f(x) is continuous on [a,b] except may be for the points x_i,

(2)

the right-limit and left-limit of f(x) at the points x_i exist.

We will say that f(x) is piecewise smooth if and only if f(x) as well as its derivatives are piecewise continuous.

Recall that $\{x_1,x_2,..,x_n\}$ is a partition of [a,b] if

a = x₁ < x₂ < ..< x_n-1 < x_n = b.

All the known results on the sum, product, and the quotient are valid for piecewise smooth functions. Except for the fundamental theorem of Calculus which needs to be modified. Indeed, we have let f(x) be piecewise smooth function on the open interval a < x < b. There is no reason for f(x) and f'(x) to be defined at the end-points a and b. But if we denote the left-limit and right-limit of f(x) at a point x₀ by

$$\lim_{x \rightarrow x_0 -} f(x) = f(x_0 -)\;\;\;\mbox{and}\;\;\lim_{x \rightarrow x_0 +} f(x) = f(x_0 +),$$

the fundamental theorem of Calculus translates into

$$\lim_{x \rightarrow b-}f(x) - \lim_{x \rightarrow a+}f(x) = \int_{a}^{b} f'(x)dx.$$

Before we state the fundamental result on convergence of Fourier series, we need some intermediary results.

Result 1. If f(x) and f'(x) are piecewise continuous on [a,b], then

$$\lim_{\lambda \rightarrow \infty} \int_{a}^{b}f(x)\sin(x\lambda) dx = 0,$$

and

$$\lim_{\lambda \rightarrow \infty} \int_{a}^{b}f(x)\cos(x\lambda) dx = 0.$$

Proof.

Remark. Recall that our initial problem is to approximate a function globally (on an interval versus Taylor approximations which are local). In this context, the approximation of f(x) will be done via the Fourier polynomials

$$f_N(x) = a_0 + \sum_{n = 1}^{n = N} (a_n \cos(nx) + b_n \sin(nx)).$$

These Fourier polynomials will be called the Fourier partial sums. Since

$$f_N(x) = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t)dt + \sum_{n = 1... ... \int_{-\pi}^{\pi} f(t)(\cos(nx)\cos(nt) + \sin(nx)\sin(nt))dt,$$

we obtain

$$f_N(x) = \frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[\frac{1}{2} + \sum_{n=1}^{n=N} \cos(n(t-x))\right]dt.$$

Set

$$D_N(\alpha) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n=1}^{n=N} \cos(n\alpha)\right).$$

We have the following result:

Result 2. We have

$$D_N(\alpha) = \frac{\sin((N+\frac{1}{2})\alpha)} {2\pi \sin(\frac{\alpha}{2})}.$$

Proof.

Using the result above, we get

$$f_N(x) = \frac{1}{\pi}\int_{-\pi}^{\pi} f(t)\frac{\sin\left(... ...t-x)\right)}{2\sin\left(\displaystyle \frac{t-x}{2}\right)} dt.$$

This formula is quite interesting since it gives the Fourier polynomials of f(x) without the coefficients.

Definition. The Dirichlet kernel is defined by

$$D_N(x) = \left\{ \begin{array}{lll} {\displaystyle \frac{\sin... ...e \frac{2N+1}{2\pi}} & x = 0, \pm 2\pi,... \end{array} \right.$$

The function D_N(x) is continuous and periodic, with $$\frac{2\pi}{N}$$ as its period. Using the formula above, we get

$$\int_{-\pi}^{0} D_N(x)dx = \int_{0}^{\pi} D_N(x)dx = \frac{1}{2}.$$

Now we are ready to state and prove the fundamental result on convergence of Fourier series, due to Dirichlet.

Theorem. Let f(x) be a function, which is twice differentiable, such that f(x), f'(x), and f''(x) are piecewise continuous on the interval $[-\pi,\pi]$. Then, for any $x \in [-\pi,\pi]$, the sequence of Fourier partial sums $\{f_N(x)\}$ converges $$\frac{1}{2}(f(x-) + f(x+))$$, as n tends to $\infty$.

Recall that the notation f(x+) (resp. f(x-)) represent the right-limit and left-limit respectively of f at the point x. Let us associate to f the new function S(f) defined by

$$S(f)(x) = \left\{ \begin{array}{cll} f(x) & \mbox{ if $f$ is... ...& \mbox{ if $f$ is discontinuous at $x$.} \end{array} \right.$$

The conclusion of the Theorem above translates into

$$\lim_{N \rightarrow \infty} f_N(x) = S(f)(x),\;\; \mbox{for any $x \in [-\pi,\pi]$}.$$

Proof.

Example. Show that

$$\frac{2}{\pi} \sum_{n=1}^{\infty} \frac{\Big(1 - (-1)^n\Big)}... ...\pi \\ 0 & x = 0 \\ -1 & -\pi < x < 0 \end{array} \right.$$

Answer. Set

$$f(x) = \left\{ \begin{array}{lll} 1 & 0 < x < \pi \\ 0 & x = 0 \\ -1 & -\pi < x < 0 . \end{array} \right.$$

This function satisfies the assumptions of the main Theorem. Before, we use the Thoerem's conclusion, let us find its Fourier series. We have

$$a_n = 0,\;\; \mbox{for any }\;\; n \geq 0.$$

Easy calculations give

$$b_n = \frac{2}{\pi} \frac{(1 - (-1)^n)}{n},\;\; \mbox{for any $n \geq 1$}.$$

The Theorem's conclusion gives the desired identity.

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