Fourier Series

Abstract

Fourier series, Dirichlet kernel, pointwise and uniform convergence of Fourier series, Fejér kernel and Cesàro summability, functions with arbitrary periods, even and odd functions, orthonormal expansions, L² convergence and Riesz–Fischer Theorem, Korovkin’s Theorem, Weierstrass’ Theorem. Example of continuous periodic function whose Fourier series diverges at a point.

Appendix

The purpose of this appendix is to prove two results. One is a weaker version of Fejér’s Theorem 4.3.4, in which continuity is assumed everywhere and which can be deduced from a theorem due to P. P. Korovkin . The latter is of independent interest and has been the subject of much research and generalisation. The second is the existence of a continuous function with a Fourier series that does not converge at 0. The construction of the function that we present is due to Fejér but du Bois-Reymond was the first to construct such a function in 1876.

Definition

Let C[a, b] be the space of all continuous real-valued functions on the interval [a, b]. A map P:C[a, b] → C[a, b] (also called an operator in C[a, b]) is said to be positive if P(f) ≥ 0 wherever f ≥ 0. It is said to be linear if $P(\upalpha\,f +\upbeta\,g) =\upalpha\,P(f) +\upbeta\,P(g)$ for all $\upalpha,\upbeta \in {\mathbb{R}}$ and all f, g ∈ C[a, b].

It may be observed that a linear positive operator P is monotone , in the sense that $f \le g \Rightarrow P(f) \le P(g)$ . Also, one can have linear positive operators in the space C_2π[− π, π] of all continuous functions on [− π, π] with f(−π) = f(π). The functions f₀ = 1, f₁ = cos and f₂ = sin belong to $C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$. Any $f \in C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$ can be uniquely extended to all of ${\mathbb{R}}$ by periodicity and the space of the extended functions is essentially the same as $C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$. In effect, we may assume every function in $C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$ to be so extended and speak of its Fourier series.

One often writes P(f) as simply Pf, especially when the operator is denoted by an upper case letter and the function to which it is applied is denoted by a lower case letter.

Theorem

(Korovkin’s Theorem) Let {P_n}_n≥1 denote a sequence of positive linear operators on $C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$. In order that P_nf → f uniformly for every $f \in C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$, it is necessary and sufficient that such convergence occur for f₀ = 1, f₁ = cos and f₂ = sin.

Proof

We first prove sufficiency. Suppose the uniform convergence occurs for the three particular functions f₀ = 1, f₁ = cos and f₂ = sin. Put

$$ \upvarphi_{y} (x) = \sin^{2} \frac{y - x}{2} = \frac{1}{2}(1 - \cos (y - x)) = \frac{1}{2}(1 - \cos \;y\;\cos \;x - \sin \;y\;\sin \;x). $$

Then we have

$$ \upvarphi_{y} = \frac{1}{2}(f_{0} - f_{1} \cos \;y - f_{2} \sin \;y) $$

and

$$ P_{n} (\upvarphi_{y} ) = \frac{1}{2}(P_{n} (f_{0} ) - P_{n} (f_{1} )\cos \;y - P_{n} (f_{2} )\sin \;y). $$

Thus

$$ \begin{aligned} (P_{n} \upvarphi_{y} )(y) & = \frac{1}{2}\left\{ {[(P_{n} f_{0} )(y) - 1] - \cos \;y[(P_{n} f_{1} )(y) - \cos \;y] - \sin \;y[(P_{n} f_{2} )(y) - \sin \;y]} \right\} \\ & \le \frac{1}{2}\left\{ {\left\| {P_{n} f_{0} - f_{0} } \right\| + \left| {\cos \;y} \right|\left\| {P_{n} f_{1} - f_{1} } \right\| + \left| {\sin \;y} \right|\left\| {P_{n} f_{2} - f_{2} } \right\|} \right\}, \\ \end{aligned} $$

where $\left\| f \right\|$ means sup $\{ |f(z)|:z \in [ - {\uppi },{\uppi }]\}$ for any $f \in C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$. Since |cos y| and |sin y| are bounded on [− π, π], it follows that $P_{n} {\upvarphi }_{y} (y)$ converges uniformly to 0 in y.

Now let f be an arbitrary element of $C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$ and let ε > 0 be given.

Since $f \in C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$ is uniformly continuous, there is a positive δ < π such that

$$ \left| {x - y} \right| < \updelta \Rightarrow \left| {f(x) - f(y)} \right| < \upvarepsilon . $$

(4.26)

Also, the function $f \in C_{{2{\uppi }}} [ - {\uppi },{\uppi }]$ is bounded and therefore there exists an M > 0 such that

$$ - M \le f(x) \le M\quad \quad \forall \,x \in [ - \uppi ,\uppi ]. $$

(4.27)

Let y be an arbitrary but fixed point of [− π, π]. If y − δ < x < y + δ, i.e., x ∈ (y − δ, y + δ), then

$$ \left| {f(x) - f(y)} \right| < \upvarepsilon $$

by (4.26) above and if δ ≤ x − y ≤ 2π − δ, i.e., x ∈ [y + δ, 2π + y − δ] then

$$ \left| {f(x) - f(y)} \right| \,\le\, 2M \,\le\, 2M\frac{{\upvarphi_{y} (x)}}{{\sin^{2} \frac{\updelta }{2}}}. $$

by (4.27) above combined with the following argument: if δ ≤ x − y ≤ 2π − δ, then $\frac{{\updelta }}{2} \le \frac{1}{2}(x - y) \le {\uppi } - \frac{{\updelta }}{2}$ and ${\upvarphi }_{y} (x) = \sin^{2} \frac{x - y}{2} \ge \sin^{2} \frac{{\updelta }}{2}$. Thus, for $x \in (y - {\updelta },2{\uppi } + y - {\updelta }] = (y - {\updelta },y + {\updelta }) \cup [y + {\updelta },2{\uppi } + y - {\updelta }]$, we have

$$ - \upvarepsilon - \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}\upvarphi_{y} (x) \,\le\, f(x) - f(y) \,\le\, \upvarepsilon + \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}\upvarphi_{y} (x); $$

that is,

$$ - \upvarepsilon f_{0} - \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}\upvarphi_{y} \, \le\, f - f(y)f_{0} \, \le\, \upvarepsilon f_{0} + \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}\upvarphi_{y} \quad \quad {\text{on }}(y - \updelta ,2\uppi + y - \updelta ]. $$

On using the periodicity of the functions involved, the above inequality is seen to hold on all of ${\mathbb{R}}$ and in particular on [−π, π]. Since the operators P_n are positive and linear, and hence monotone, we have

$$ \begin{aligned} - \upvarepsilon (P_{n} f_{0} )(y) - \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}(P_{n} \upvarphi_{y} )(y) & \le (P_{n} f)(y) - f(y)(P_{n} f_{0} )(y) \\ & \le \upvarepsilon (P_{n} f_{0} )(y) + \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}(P_{n} \upvarphi_{y} )(y), \\ \end{aligned} $$

which implies

$$ \left| {(P_{n} f)(y) - f(y)(P_{n} f_{0} )(y)} \right| \le \upvarepsilon \left\| {P_{n} f_{0} } \right\| + \frac{2M}{{\sin^{2} \frac{\updelta }{2}}}(P_{n} \upvarphi_{y} )(y). $$

Since P_nf₀ → f₀ and $\left( {P_{n} {\upvarphi }_{y} } \right)(y) \to 0$ uniformly in y, it follows that (P_nf) (y) → f(y) uniformly in y. This completes the proof of sufficiency.

The necessity is trivial. $\square$

Theorem

(Fejér ) The Cesàro means of the Fourier series of a continuous function of period 2π converge uniformly to the function.

Proof

It is clear from Remark 4.3.2 that the Fejér sums (operators) σ_n are positive and linear. We may complete the proof by verifying that ${\upsigma }_{n} f \to f$ uniformly when f = 1 or cos or sin. This requires only some simple computation:

$$ \begin{array}{*{20}c} {\upsigma_{n} 1 = \frac{1}{n}(1 + 1 + \cdots + 1) = 1 \to 1;} \\ {(\upsigma_{n} \cos )(x) = \frac{1}{n}(0 + \cos \;x + \cos \;x + \cdots + \cos \;x) = \frac{n - 1}{n}\cos \;x \to \;\cos \;x;} \\ {(\upsigma_{n} \sin )(x) = \frac{1}{n}(0 + \sin \;x + \sin \;x + \cdots + \sin \;x) = \frac{n - 1}{n}\sin \;x \to \;\sin \;x.\quad \quad \square } \\ \end{array} $$

Theorem

There exists a continuous function with period 2π for which the Fourier series diverges to ∞ at 0.

Proof

For integers m > n > 0, set

$$ T(x,m,n) = 2\sin \;mx\sum\limits_{k = 1}^{n} {\frac{\sin \,kx}{k}} . $$

(4.28)

By Proposition 4.2.9, the sum on the right-hand side is bounded in absolute value (by 1 + π) on ${\mathbb{R}}$ and therefore so is T(x, m, n). Using elementary trigonometric identities, we can rewrite T(x, m, n) as the sum of the sums

$$ \frac{\cos (m - n)x}{n} + \frac{\cos (m - n + 1)x}{n - 1} + \cdots + \frac{\cos (m - 1)x}{1} $$

(4.29)

and

$$ - \left[ {\frac{\cos (m + 1)x}{1} + \frac{\cos (m + 2)x}{2} + \cdots + \frac{\cos (m + n)x}{n}} \right]. $$

(4.30)

It will be convenient to refer to the sum of (4.29) and (4.30) as the expansion of T(x, m, n). Observe that the multiples of x in (4.29) and in (4.30), taken together, are in increasing order, and consequently, the expansion of T(x, m, n) consists of terms of a trigonometric series in their proper order.

Now let {n_k} and {m_k} be sequences of positive integers such that n_k < m_k and let $\sum {\upalpha_{k} }$ be a convergent series of positive terms. As T(x, m, n) is bounded in absolute value, the series

$$ \sum\limits_{k = 1}^{\infty } {\upalpha_{k} T(x,m_{k} ,n_{k} )} $$

(4.31)

converges uniformly and absolutely on ${\mathbb{R}}$. Therefore its sum is a continuous function f(x) on ${\mathbb{R}}$ with period 2π. Each T is an even function and therefore the same is true of f.

We can choose the sequences {n_k} and {m_k} such that

$$ m_{k} + n_{k} < m_{k + 1} - n_{k + 1} \quad \quad \forall \,k. $$

To do so, all we need to arrange for is that n_k+1 > 3n_k and m_k = 2n_k, so that m_k + n_k = 3n_k < n_k+1 = 2n_k+1 − n_k+1 = m_k+1 − n_k+1. This ensures that, when k > j, all the multiples of x in the expansion of T(x, m_k, n_k) are higher multiples than all those in the expansion of T(x, m_j, n_j). Therefore in view of the observation above, upon writing out the expansion of each T(x, m_k, n_k) in the sum (4.31), we get a trigonometric series with terms in the proper order. Denote the series by $\sum {a_{j} } \cos \;jx$. The partial sums of (4.31) form a subsequence of the partial sums of the trigonometric series. Since the former converge uniformly to f(x), it follows [see Problem 4.1.P7] that the Fourier series corresponding to f is in fact Σa_j cos jx.

Now consider the partial sum $s_{m_p}(x)$ of the Fourier series. [This will not be a partial sum of (4.31) because it stops midway in T(x, m_p, n_p).]

$$ \begin{aligned} s_{{m_{p} }} (x) & = \sum\limits_{j = 1}^{{m_{p} }} {a_{j} \cos \;jx} \\ & = \sum\limits_{k = 1}^{p - 1} {\upalpha_{k} T(x,m_{k} ,n_{k} )} + \upalpha_{p} \left[\frac{{\cos (m_{p} - n_{p} )x}}{{n_{p} }} \right.\\ &\left. \quad + \frac{{\cos (m_{p} - n_{p} + 1)x}}{{n_{p} - 1}} + \cdots + \frac{{\cos (m_{p} - 1)x}}{1}\right]. \\ \end{aligned} $$

Since T(0, m, n) = 0 by (4.28), we have

$$ s_{{m_{p} }} (0)=\upalpha_{p} \sum\limits_{k = 1}^{{n_{p} }} {\frac{1}{k}} . $$

By Problem 4.2.P5, $\sum\limits_{k = 1}^{{n_{p} }} {\frac{1}{k}} >$ ln n_p and therefore $s_{m_p}(0)< \upalpha_{p}$ ln n_p. If we now show that the sequences {n_k}, {m_k} and {α_k} can be so chosen that α_p ln n_p→ ∞ as p → ∞, then it will follow that the partial sums of the Fourier series of f diverge to ∞ when x = 0.

To choose {n_k}, {m_k} and {α_k} in the required manner, let α_k = 1/k² and arrange not only that m_k = 2n_k and n_k+1 > 3n_k as mentioned earlier, but also that n_k+1 > exp(k + 1)³. $\square$