Notes on Fourier analysis (part 2)

1.3. Uniform convergence

Continuity is not necessarily preserved under pointwise limits. For example, Suppose that ${f_n:[0,1]\rightarrow \mathbb{R}}$ is defined by ${f_n(x)=x^n}$ . If ${0<x<1}$ , then ${x^n\rightarrow0}$ as ${n\rightarrow\infty}$ . If ${x=1}$ , then ${x^n\rightarrow1}$ as ${n\rightarrow\infty}$ . Thus, ${f_n}$ is continuous on ${[0,1]}$ but the pointwise limit ${f}$ is not (it is discontinuous at 1). On the other hand, uniform convergence of continuous functions does guarantee continuity.

Definition 3.1 Uniform convergence The series of continuous function ${\sum^{\infty}_{n=1} f_n}$ converges uniformly on the closed interval ${[a,b]}$ if the following two conditions are satisfied: First, the series converges to the sum ${S(x)}$ for each ${x}$ value in the interval; Second, given any number ${\epsilon >0}$ , there exists a ${N\in\mathbb{N}}$ such that for all ${m\geq N}$ the partial sums ${S_m=\sum^{m}_{n=1} f_n}$ satisfy ${\mid S(x)-S_m(x) \mid\leq\epsilon}$ for every ${x}$ value in the interval.

The following theorem gives an easily applied condition for determining uniform convergence. It is especially useful in its applicaion to Fourier series.

Theorem 3.2 Weierstrass’ ${M}$ -Test If for all points ${x}$ in ${[a,b]}$ we have ${\mid f_n(x)\mid \leq M_n }$ from a certain ${n=k}$ on, and the series of positive numbers ${\sum^{\infty}_{n=k}M_n}$ converges, then the series ${\sum^{\infty}_{n=1}f_n}$ converges uniformly on ${[a,b]}$ .

Proof For each value ${x}$ in the interval ${[a,b]}$ , since ${\mid f_n(x)\mid \leq M_n }$ and ${\sum^{\infty}_{n=k}M_n}$ converges, then by comparison test, the series ${\sum^{\infty}_{n=k}f_n}$ converges absolutely. Therefore, the series ${\sum^{\infty}_{n=k}f_n}$ converges absolutely for every ${x}$ on the interval ${[a,b]}$ .

Given some ${\epsilon>0}$ , then there exist an ${ \in \mathbb{N}}$ such that ${m>N}$ implies ${\sum^{\infty}_{n=m+1}M_n < \epsilon}$ .

By Triangle inequality, we have for each ${x}$ in ${[a,b]}$

$\displaystyle \lvert S(x)-S_m(x)\rvert \leq \lvert\sum^{\infty}_{n=m+1}f_n(x)\rvert \leq\sum^{\infty}_{n=m+1}\lvert f_n(x)\rvert \leq\sum^{\infty}_{n=m+1}M_n < \epsilon \ \ \ \ \ (62)$

Thus, for all ${m\geq\mathbb{N}}$ ,

$\displaystyle \mid S(x)-S_m(x) \mid < \epsilon \ \ \ \ \ (63)$

for every ${x}$ in ${[a,b]}$ .

${\square}$

In the case of Fourier series, we have the following estimates

$\displaystyle \hspace{0.5cm} \lvert a_n\mathrm{cos}n\theta\rvert\leq \lvert a_n \rvert \hspace{0.5cm} \lvert b_n\mathrm{sin}n\theta\rvert\leq \lvert b_n \rvert \hspace{0.5cm} \lvert c_ne^{in\theta}\rvert\leq \lvert c_n \rvert \ \ \ \ \ (64)$

Hence the Weierstrass’s ${M}$ -test will apply to a Fourier series in trigonometric form if ${\sum^{\infty}_0\lvert a_n \rvert<\infty}$ and ${\sum^{\infty}_1\lvert b_n \rvert<\infty}$ , and to Fourier series in exponential form if ${\sum^{\infty}_{-\infty}\lvert c_n \rvert<\infty}$ .

Another theorem that we will use for the proof of uniform convergence is the Cauchy-Schwarz Inequality.

Theorem 3.3 Cauchy-Schwarz Inequality Let ${ {a_n}^N_{n=1}}$ and ${ {b_n}^N_{n=1}}$ be two finite sets of real numbers. Then,

$\displaystyle (\sum^{N}_{n=1}a_nb_n)^2\leq (\sum^{N}_{n=1}a_n^2)(\sum^{N}_{n=1}b_n^2) \ \ \ \ \ (65)$

When expressed in vector form, the Cauchy-Schwarz inequality says that the dot product of two vectors is bounded by the product of their norms.

Proof Expanding out the brackets and collecting identical terms, we have

$\displaystyle \begin{aligned} \sum^{N}_{i=1}\sum^{N}_{j=1}(a_ib_j-a_jb_i)^2&=\sum^{N}_{i=1}a_i^2\sum^{N}_{j=1}b_j^2+\sum^{N}_{j=1}a_j^2\sum^{N}_{i=1}b_i^2-2\sum^{N}_{i=1}a_ib_i\sum^{N}_{j=1}a_jb_j\\ &= 2(\sum^{N}_{i=1}a_i^2)(\sum^{N}_{i=1}b_i^2)-2 (\sum^{N}_{i=1}a_ib_i)^2 \end{aligned} \ \ \ \ \ (66)$

The left-hand side of the equation is greater than or equal to zero since it is a sum of the squares of real numbers. Thus,

$\displaystyle (\sum^{N}_{i=1}a_i^2)(\sum^{N}_{i=1}b_i^2)\geq(\sum^{N}_{i=1}a_ib_i)^2 \ \ \ \ \ (67)$

${\square}$

The graph of periodic functions that are both continuous and piecewise smooth is a smooth curve except that is can have “corners” where the derivatives jump. The fundamental theorem of calculus,

$\displaystyle f(b)-f(a)=\int^b_a f'(\theta) \mathrm{d}\theta \ \ \ \ \ (68)$

applies to functions ${f}$ that are continuous and piecewise smooth, even though ${f'}$ is defined at the “corners”. To see this, let ${f}$ to be differentiable except at point ${c\in(a,b)}$ , we have

$\displaystyle \begin{aligned} \int^b_a f'(\theta) \mathrm{d}\theta &= \int^b_c f'(\theta) \mathrm{d}\theta+\int^c_a f'(\theta) \mathrm{d}\theta\\ &= (f(b)-f(c))+(f(c)-f(a))\\ &= f(b)-f(a) \end{aligned} \ \ \ \ \ (69)$

We introduce the following theorem as a preliminary step towards the main convergence theorem.

Theorem 3.4 Suppose ${f}$ is ${2\pi}$ -periodic, continuous, and piecewise smooth. Let ${a_n}$ , ${b_n}$ , and ${c_n}$ be the Fourier coefficients of ${f}$ defined in (2.5) and (2.6), and let ${a_n'}$ , ${b_n'}$ , and ${c_n'}$ be the corresponding Fourier coefficients of ${f'}$ . Then

$\displaystyle \hspace{0.5cm} a_n'=nb_n, \hspace{0.5cm} b_n'=-na_n, \hspace{0.5cm} c'_n=inc_n \ \ \ \ \ (70)$

Proof This is a simple matter of integration by parts.

${\square}$

From theorem 3.4 we obtain the following results on differentiation and integration of Fourier series.

Theorem 3.5 Suppose ${f}$ is ${2\pi}$ -periodic, continuous, and piecewise smooth, and suppose also that ${f'}$ is piecewise smooth. If

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\mathrm{cos}(n\theta)+b_n\mathrm{sin}(n\theta))=\sum_{-\infty}^{\infty}c_ne^{in\theta} \ \ \ \ \ (71)$

is the Fourier series of ${f(\theta)}$ , then ${f'(\theta)}$ is the sum of the derived series

$\displaystyle \frac{1}{2}a_0+\sum_{n=1}^{\infty}(nb_n\mathrm{cos}(n\theta)-na_n\mathrm{sin}(n\theta))=\sum_{-\infty}^{\infty}inc_ne^{in\theta} \ \ \ \ \ (72)$

for all ${\theta}$ at which ${f'(\theta)}$ exists. At the exceptional points ${f'}$ has jumps, the series converges to ${\frac{1}{2}[f'(\theta_-)+f'(\theta_+)]}$ .

Proof Since ${f'}$ is piecewise smooth, by Theorem 2.6, it is the sum of it Fourier series at every point. By theorem 3.4, the coefficients of ${e^{in\theta}}$ , ${\mathrm{sin}(n\theta)}$ , ${\mathrm{cos}(n\theta)}$ in this series are ${inc_n}$ , ${-na_n}$ , and ${nb_n}$ respectively. Thus theorem 3.5 follows.

${\square}$

Theorem 3.6 Suppose ${f}$ is ${2\pi-}$ periodic and piecewise continuous, with Fourier coefficients ${a_n}$ , ${b_n}$ , ${c_n}$ , and let ${F(\theta)=\int^\theta_{0}f(\phi)\mathrm{d}\phi}$ . If ${c_0(=\frac{1}{2}a_0)=0}$ , then for all ${\theta}$ we have

$\displaystyle F(\theta)=\frac{1}{2}A_0+\sum_{n=1}^{\infty}(\frac{a_n}{n}\mathrm{sin}(n\theta)-\frac{b_n}{n}\mathrm{cos}(n\theta))=C_0+\sum_{n \neq 0}\frac{c_n}{in}e^{in\theta} \ \ \ \ \ (73)$

where the constant term is the mean value of ${F}$ on ${[-\pi,\pi]}$ :

$\displaystyle C_0=\frac{1}{2}A_0=\frac{1}{2\pi}\int^{\pi}_{-\pi}F(\theta)\mathrm{d}\theta \ \ \ \ \ (74)$

The series on the right of (72) is the series obtained by formally integrating the Fourier series of ${f}$ term by term, whether the latter series actually converges or not. If ${c_0 \neq 0}$ , the sum of the series on the right of (72) is ${F(\theta)-c_0\theta}$ .

Proof ${F}$ is continuous and piecewise smooth, being the integral of a piecewise continuous function. Moreover, if ${c_0=0}$ , ${F}$ is ${2\pi-}$ periodic, for

$\displaystyle F(\theta+2\pi)-F(\theta)=\int^{\theta+2\pi}_{\theta} f(\phi)\mathrm{d}\phi=\int^{\pi}_{-\pi} f(\phi)\mathrm{d}\phi=2\pi c_0=0 \ \ \ \ \ (75)$

Hence, by theorem 2.6, ${F(\theta)}$ is the sum of its Fourier series at every ${\theta}$ . But by theorem 3.4 applied to F, the Fourier coefficients ${A_n, B_n}$ , and ${C_n}$ of ${F}$ are related to those of ${f}$ by

$\displaystyle \hspace{0.5cm} A_n'=-\frac{b_n}{n}, \hspace{0.5cm} B_n'=\frac{a_n}{n}, \hspace{0.5cm}C_n=\frac{c_n}{in} \hspace{0.5cm} (n\neq0). \ \ \ \ \ (76)$

The formula (74) for the constant ${C_0}$ or ${\frac{1}{2}A_0}$ is just the usual formula for the zeroth Fourier coefficient of ${F}$ . If ${c\neq0}$ , these arguments can be applied to the function ${f(\theta)-c_0}$ rather than ${f(\theta)}$ , yielding the final assertion.

${\square}$

We will now prove a theorem on uniform convergence of Fourier series.

Theorem 3.7 If ${f}$ is ${2\pi-}$ periodic, continuous, and piecewise smooth, then the Fourier series of ${f}$ converges to ${f}$ absolutely and uniformly on ${\mathbb{R}}$ .

Proof Let ${c_n'}$ denote the Fourier coefficients of ${f'}$ . By theorem 3.4 we know that ${c_n=(in)^-1c_n'}$ for ${n \neq 0}$ , and by (29) (Bessel’s inequality) applied to ${f'}$ ,

$\displaystyle \sum^{\infty}_{-\infty}|c_n'|^2\leq \frac{1}{2\pi}\int^{\pi}_{-\pi}|f'(\theta)|^2\mathrm{d}\theta<\infty \ \ \ \ \ (77)$

Hence, by the Cauchy-Schwartz inequality,

$\displaystyle \sum^{\infty}_{-\infty}|c_n|=\lvert c_0\rvert+\sum_{n\neq0}\lvert \frac{c_n'}{in} \rvert\leq \lvert c_0\rvert+(\sum_{n\neq 0}\frac{1}{n^2})^{\frac{1}{2}}(\sum_{n\neq 0}\lvert c_n'\rvert^2)^{\frac{1}{2}} <\infty \ \ \ \ \ (78)$

since ${\sum_{n\neq 0}(\frac{1}{n^2})=2\sum_{n=1}^{\infty}(\frac{1}{n^2})<\infty}$ . This completes the proof.

${\square}$

Notes on Fourier analysis (part 2)

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