Unravel the Power of Trigonometry: The Weierstrass Approximation Theorem Revealed

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\[\frac{1}{\pi}\int_{-\pi}^{\pi}[f(x)-s_{n}(f)]^{2}dx=\frac{1}{\pi}\int_{-\pi}^{ \pi}f(x)^{2}\,dx-\frac{a_{0}^{2}}{2}-\sum_{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\]

\[=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)^{2}\,dx-\frac{1}{\pi}\int_{-\pi}^{\pi}s_{n }(f)(x)^{2}\,dx.\qed\]

**Corollary 19** (The Weierstrass approximation theorem- trigonometric form).: _Let \(f(x)\) be continuous function on the interval \([-\pi,\pi]\) and \(f(-\pi)=f(\pi)\). Then for each \(\epsilon>0\) there is a trigonometric polynomial \(T(x)\) such that_

\[|f(x)-T(x)|<\epsilon\quad\text{for}\quad-\pi\leq x\leq\pi.\]

The proof of this theorem is left as an exercise (see Exercise 6 below). Recall that a function \(f:[a,b]\to\mathbb{R}\) is called piecewise continuous if \([a,b]\) has a finite partition \(a=x_{0}

* Every continuous periodic function can be approximated uniformly by a continuous piecewise linear function with the same period.

* The Fourier coefficients of a continuous, periodic, piecewise linear function \(f\) satisfy \(|a_{n}|\leq\frac{C}{n^{2}}\) and \(|b_{n}|\leq\frac{C}{n^{2}}\) for some constant \(C>0\). This shows that the Fourier series converge uniformly to \(f(x)\).

Observe that the trigonometric form of the Weierstrass Approximation Theorem is closely related to the algebraic form, as given in the Section 2.4 on approximation by polynomials in this book.

**Remark 57**.: Assume \(f\) is a real-valued Riemann integrable function and its \(L^{2}\)-norm is

\[||f||_{2}=\left(\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)^{2}\,dx\right)^{1/2}.\]

Let \(\mathcal{T}_{n}\) denote the collection of trigonometric polynomials of degree at most \(n\). Then we can restate the mean-square theorem above as: the partial sum \(s_{n}(f)\) is the nearest point to \(f\) out of \(\mathcal{T}_{n}\), that is

\[\inf_{T\in\mathcal{T}_{n}}||f-T||_{2}=||f-s_{n}(f)||_{2}.\]

Following Bessel’s inequality is valid for more general systems of orthogonal functions. Here our aim is to show that the Fourier coefficients \(a_{k}\) and \(b_{k}\) of square integrable functions tend to zero as \(n\to\infty\).

**Corollary 20** (Bessel’s inequality).: _Let \(s_{n}(f)\) be the \(n\)th partial sum of the Fourier series for \(f\); then_

\[||s_{n}(f)||_{2}\leq||f||_{2}.\]

Proof

: Note that

\[||f-s_{n}(f)||_{2}^{2}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)^{2}\,dx-\frac{a_{0}^{ 2}}{2}-\sum_{k=1}^{n}(a_{k}^{2}+b_{k}^{2})=||f||_{2}^{2}-||s_{n}(f)||_{2}^{2}.\]

Since \(||f-s_{n}(f)||_{2}\geq 0\), we have

\[||s_{n}(f)||_{2}^{2}=\int_{-\pi}^{\pi}s_{n}(f)(x)^{2}\,dx=\frac{a_{0}^{2}}{2}+ \sum_{k=1}^{n}(a_{k}^{2}+b_{k}^{2})\leq\int_{-\pi}^{\pi}f(x)^{2}\,dx=||f||_{2} ^{2}.\qed\]

**Remark 58** (Uniform Convergence).: The trigonometric polynomials belong to the set of all \(2\pi\)-periodic continuous functions \(f:\mathbb{R}\to\mathbb{R}\), a space we will denote by \(C(2\pi)\). If the Fourier series of a function \(f\in C(2\pi)\) is uniformly convergent, then the series must actually converge to \(f\). Clearly, if a trigonometric series is uniformly convergent then its sum defines a continuous function (recall our discussion of uniform convergence on “modes of convergence” Section 2.3). Let us call this function \(g\in C(2\pi)\) in this case. Now, all we need is to observe that \(g\) has the same Fourier coefficients as \(f\). This is so because if \(s_{n}(f)\) converges uniformly to \(g\), then \(s_{n}(f)\cos kx\) converges uniformly to \(g(x)\cos kx\). Interchanging the limit and integral, we obtain

\[\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos kx\,dx=\lim_{n\to\infty}\frac{1}{\pi} \int_{-\pi}^{\pi}s_{n}(f)(x)\cos kx\,dx=a_{k}.\]Similarly, \(\frac{1}{\pi}\int_{-\pi}^{\pi}g(x)\cos kx\,dx=b_{k}\).

Moreover, if the Fourier coefficients for \(f\) satisfy

\[\sum_{n}|a_{n}|<\infty\quad\text{and}\quad\sum_{n}|b_{n}|<\infty,\]

then using the well-known Weierstrass M-test (see Theorem 97 in Section 2.3), we can conclude that the Fourier series for \(f\) is uniformly convergent in \(\mathbb{R}\). Therefore, if we are given that \(f\in C(2\pi)\), then by the above discussion it follows that the Fourier series for \(f\) converges uniformly to \(f\).

In the following we study partial sums of an Fourier series via the Dirichlet kernel. To better understand the pointwise convergence of Fourier series, it would be useful to have an expression for the partial sums \(s_{n}(f)\) not involving a sum; this is done with the Dirichlet formula.

**Definition 83** (Dirichlet Kernel).: The function

\[D_{n}(x)=\frac{\sin(n+\frac{1}{2})x}{2\sin\frac{1}{2}x}=\frac{1}{2}+\sum_{k=1 }^{n}\cos kx\]

is known as the _Dirichlet kernel_ (Figure 2.18).

First note that the equality of

\[\frac{\sin(n+\frac{1}{2})x}{2\sin\frac{1}{2}x}=\frac{1}{2}+\sum_{k=1}^{n}\cos kx\]

follows from the identity

\[2\sin\frac{1}{2}x\cos kx=\sin\left(k+\frac{1}{2}\right)x-\sin\left(k-\frac{1}{ 2}\right)x.\]

**Lemma 11**.: _The following are the properties of the Dirichlet kernel._

Figure 2.18: The Dirichlet kernel for \(n=10\)

* _The Dirichlet kernel is an even function with period_ \(2\pi\)_, i.e.,_ \(D_{n}(-x)=D_{n}(x)\) _and_ \(D_{n}(x+2\pi)=D_{n}(x)\)_._
* \(\frac{1}{\pi}\int_{-\pi}^{\pi}D_{n}(x)\,dx=1\quad n=1,2,\dots\)_._
* _The_ \(n\)_th partial sum_ \(s_{n}(x)\) _of the Fourier series of a function_ \(f\) _can be written as_ \[s_{n}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}D_{n}(f)f(x+t)\,dt.\] _The above formula is called the Dirichlet formula for the_ \(n\)_th partial sum of a Fourier series._
* \(\int_{-\pi}^{\pi}|D_{n}(t)|\,dt\geq C\log n,\quad n\to\infty\) _for some constant_ \(C>0\)_._

Proof

: a) is clear and b) follows from the simple fact that

\[\int_{-\pi}^{\pi}\cos kxdx=0\,\,\,\text{for $k=1,2,\dots$}.\]

For c) we write:

\[s_{n}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}\left\{\frac{1}{2}+\sum_{k=1}^{n}(\cos k x \cos kt+\sin ka\sin kt)\right\}f(t)dt\]

and therefore,

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

If \(f\) is \(2\pi\)-periodic, then we can also write

\[s_{n}(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}D_{n}(t)f(x+t)dt\]

and moreover

\[s_{n}(x)-f(x)=\frac{1}{\pi}\int_{-\pi}^{\pi}D_{n}(t)[f(x+t)-f(x)]dt.\]

For part d) we refer the reader to [49], p. 50.

In light of the Dirichlet formula we see that in order to prove \(s_{n}(x)\to f(x)\) as \(n\to\infty\), we only need to show that the above integral tends to zero. However, \(D_{n}(t)\) alternates sign in \([-\pi,\pi]\) and the property d) in the above lemma makes it clear that we need a better kernel; this “good” kernel is called the Fejer kernel for approximation, as we shall discuss soon. Nevertheless, one can find extra conditions on the function \(f\) under which \(s_{n}\to f\) uniformly (see [49], p. 41).

Even though the Fourier coefficients \(a_{k}\) and \(b_{k}\) are determined by the values of the function \(f\) throughout the interval \([-\pi,\pi]\), the convergence of the resulting Fourier series at any point \(x\) depends on the behavior of \(f\) in an arbitrarily small neighborhood of \(x\) as illustrated in the following theorem.

**Theorem 108** ([20], p. 205).: _Let \(f\) be square integrable over \([-\pi,\pi]\), and let \(s_{n}\) be the \(n\)th partial sum of its Fourier series. Let \(f\) be extended to the real line with \(f(x+2\pi)=f(x)\). Suppose at some point \(x\in\mathbb{R}\) the extended function satisfies the condition_

\[|f(x+t)-f(x)|\leq C|t|,\quad\text{where}\quad|t|<\delta,\]