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Trigonometric Series

  1. Let \(T\) be the unit circle in the complex plane, i.e., the set of all complex numbers of absolute value \(1\). If \(F\) is any function on \(T\) and if \(f\) is defined on \(R^1\) by \[f(t)=F(e^{it})\] Then \(f\) is a periodic function of period \(2\pi\). Conversely, if \(f\) is a function on \(R^1\), with period \(2\pi\), then there is a function \(F\) on \(T\) such that \[f(t)=F(e^{it})\] holds.
    We define \(L^p(T)\), for \(1\leq p<\infty\), to be the class of all complex, Lebesgue measurable, \(2\pi\)-periodic functions on \(R^1\) for which the norm \[\lVert f\rVert_p=\Biggl\{\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(t)|^pdt\Biggr\}^{1/p}\] is finite. In other words, we are looking at \(L^p(\mu)\), where \(\mu\) is Lebesgue measure on \([0,2\pi]\), divided by \(2\pi\). \(L^{\infty}(\mu)\) will be the class of all \(2\pi\)-periodic members of \(L^{\infty}(R^1)\), with the essential supremum norm, and \(C(T)\) consists of all continuous complex functions on \(T\) (or equivalently, of all continuous complex, \(2\pi\)-periodic functions on \(R^1\)), with norm \[\lVert f\rVert_{\infty}=\sup_{t}|f(t)|\]

  2. A trigonometric polynomial is a finite sum of the form: \[f(t)=a_0+\sum_{n=1}^{N}(a_n\cos nt+b_n\sin nt),\quad(t\in R^1)\] where \(a_0,a_1,\cdots,a_N\) and \(b_0,b_1,\cdots,b_N\) are complex numbers and since \[\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})\] \[\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})\] \[e^{inx}=\cos(nx)+i\sin(nx)\] then \[f(t)=a_0+\sum_{n=1}^{N}(a_n\cos nt+b_n\sin nt),\quad(t\in R^1)\\ =a_0+\sum_{n=1}^{N}\Biggl(a_n\frac{1}{2}(e^{int}+e^{-int})+b_n\frac{1}{2i}(e^{int}-e^{-int})\Biggr)\\ =\sum_{n=-N}^{N}c_ne^{int}\] It is clear that every trigonometric polynomial has period \(2\pi\).

  3. \(Z\) is the set of all integers. Put \(u_n(t)=e^{int},\quad(n\in Z)\) If we define the inner product in \(L^2(T)\) by \[(f,g)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt\] then \[(u_n,u_m)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(n-m)t}dt=\begin{cases} 1&\text{if } n=m\\ 0&\text{if } n\ne m \end{cases}\] Thus \(\{u_n:n\in Z\}\) is an orthonormal set in \(L^2(T)\), usually called the trigonometric system.

  4. The maximality (or completeness) of the trigonometric system will be proved as soon as we can show that the set of all trigonometric polynomials is dense in \(L^2(T)\). \(C(T)\) consists of all continuous complex functions on \(T\). Since \(C(T)\) is dense in \(L^2(T)\), (note that \(T\) is compact), \(\Biggl(\) For \(1\leq p<\infty\), \(C_c(X)\) is dense in \(L^p(\mu)\).
    Define \(S\) as in last Theorem. If \(s\in S, \epsilon>0\), there exists a \(g\in C_c(X)\) such that \(g(x)=s(x)\) except on a set of measure \(<\epsilon\), and \(|g|\leq \lVert s\rVert_{\infty}\) (Lusin’s theorem: An almost-everywhere finite function is measurable if and only if it is a continuous function on nearly all its domain). Hence \[\lVert g-s\rVert_{p}\leq\lVert g\rVert_{p}+\lVert s\rVert_{p}\leq2\epsilon^{1/p}\lVert s\rVert_{\infty}\] Since \(S\) is dense in \(L^p(\mu)\), this completes the proof. If \(1\leq p<\infty\), \(C_c(R^k)\) is dense in \(L^p(R^k)\) and \(L^p(R^k)\) is complete. Thus \(L^p(R^k)\) is the completion of the metric space which is obtained by endowing \(C_c(R^k)\) with the \(L^p\)-metric. \(\Biggl)\) it suffices to show that to every \(f\in C(T)\) and to every \(\epsilon>0\) there is a trigonometric polynomial \(P\) such that \[\lVert f-P\rVert_{2}<\epsilon\]
    Since \(\lVert g\rVert_{2}\leq\lVert g\rVert_{\infty}\) for every \(g\in C(T)\), the estimate \(\lVert f-P\rVert_{2}\leq\epsilon\) will follow from \(\lVert f-P\rVert_{\infty}\leq\epsilon\), and it is this estimate which we shall prove.
    Suppose we had trigonometric polynomials \(Q_1,Q_2,Q_3.\cdots\), with the following properties:

  1. \[Q_k(t)\ge0,\quad t\in R^1\]
  2. \[\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_k(t)dt=1\]
  3. If \(\eta_k(\delta)=\sup\;\{Q_k(t):\delta\leq|t|\leq\pi\}\), then \[\lim_{k\to\infty}\eta_k(\delta)=0,\quad \forall \delta>0\]
    To each \(f\in C(T)\) we associate the functions \(P_k\) defined by \[P_k(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t-s)Q_k(s)ds\quad(k=1,2,3,\cdots)\] If we replace \(s\) by \(-s\) and then by \(s-t\), the periodicity of \(f\) and \(Q_k\) shows that the value of the integral is not affected. Hence \[P_k(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(s)Q_k(t-s)ds\quad(k=1,2,3,\cdots)\] Since each \(Q_k\) is a trigonometric polynomial, \(Q_k\) is the form \[Q_k(t)=\sum_{n=-N_k}^{N_k}a_{n,k}e^{int}\] and if we replace \(t\) by \(t-s\) then \[Q_k(t-s)=\sum_{n=-N_k}^{N_k}a_{n,k}e^{in(t-s)}\] then \[P_k(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(s)Q_k(t-s)ds\quad(k=1,2,3,\cdots)\\ =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(s)\Biggl(\sum_{n=-N_k}^{N_k}a_{n,k}e^{in(t-s)}\Biggr)ds\] we see that each \(P_k\) is a trigonometric polynomial.
    For \(\epsilon>0\), since \(f\) is uniformly continuous on \(T\), there exists a \(\delta>0\) such that \(|f(t)-f(x)|<\epsilon\) whenever \(|t-s|<\delta\), since \[P_k(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t-s)Q_k(s)ds\quad(k=1,2,3,\cdots)\] and (b) \[\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_k(t)dt=1\] we have \[P_k(t)-f(t)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\Bigl[f(t-s)-f(t)\Bigr]Q_k(s)ds\] and (a) \[Q_k(t)\ge0,\quad t\in R^1\] implies, for all \(t\), that \[|P_k(t)-f(t)|\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(t-s)-f(t)|Q_k(s)ds=A_1+A_2\] where \(A_1\) is the integral over \([-\delta,\delta]\) and \(A_2\) is the integral over \([-\pi,-\delta]\cup[\delta,\pi]\). In \(A_1\), the integrand is less than \(\epsilon Q_k(s)\), so \(A_1<\epsilon\). In \(A_2\), we have \(Q_k(s)\leq\eta_k(\delta)\), hence \[A_2\leq2\lVert f\rVert_{\infty}\cdot \eta_k(\delta)<\epsilon\] for sufficiently large \(k\), by (c) If \(\eta_k(\delta)=\sup\;\{Q_k(t):\delta\leq|t|\leq\pi\}\), then \[\lim_{k\to\infty}\eta_k(\delta)=0,\quad \forall \delta>0\]
    Since these estimates are independent of \(t\), we have proved that \[\lim_{k\to\infty}\lVert f-P_k\rVert_{\infty}=0\]
    It remains to construct the \(Q_k\). Put \[Q_k(t)=c_k\Bigl(\frac{1+\cos t}{2}\Bigr)^k\] where \(c_k\) is chosen so that (b) \[\frac{1}{2\pi}\int_{-\pi}^{\pi}Q_k(t)dt=1\] holds. Since (a) \(Q_k(t)\ge0,\quad t\in R^1\) is clear, we only need to show (c).
    Since \(Q_k\) is decreasing on \([0,\pi]\), it follows that \[Q_k(t)\leq Q_k(\delta)\leq\frac{\pi(k+1)}{2}\Bigl(\frac{1+\cos \delta}{2}\Bigr)^k,\quad(0<\delta\leq|t|\leq\pi)\] This implies (c), since \(1+\cos\delta<2, 0<\delta\leq\pi\).
  1. If \(f\in C(T)\) and \(\epsilon>0\), there is a trigonometric polynomial \(P\) such that \[|f(t)-P(t)|<\epsilon,\forall\text{ real }t\]

  2. For any \(f\in L^1(T)\), we define the Fourier coefficients of \(f\) by the formula \[\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\quad(n\in Z)\] where, we recall, \(Z\) is the set of all integers. We thus associate with each \(f\in L^1(T)\) a function \(\hat{f}\) on \(Z\). The Fourier series of \(f\) is \[\sum_{-\infty}^{\infty}\hat{f}(n)e^{int}\] and its partial sums are \[s_N(t)=\sum_{-N}^{N}\hat{f}(n)e^{int}\quad(N=0,1,2,\cdots)\]
    Since \(L^2(T)\subset L^1(T)\), \[\hat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\quad(n\in Z)\] can be applied to every \(f\in L^2(T)\).

  3. Riesz-Fischer theorem: Let \(u_a:a\in A\) be an orthonormal set in \(H\), and let \(P\) be the space of all finite linear combinations of the vectors \(u_a\). The inequality \[\sum_{a\in A}|\hat{x}(a)|^2\leq\lVert x\rVert^2,\quad \forall x \in H\] holds and \(x\to\hat{x}\) is a continuous linear mapping of \(H\) onto \(\ell^2(A)\) whose restriction to the closure \(\bar{P}\) of \(P\) is an isometry of \(\bar{P}\) onto \(\ell^2(A)\).
    Let \(F\) be a finite subset of \(A\), let \(M_F\) be the span of \(\{u_a:a\in F\}\), then \[\sum_{a\in F}|\hat{x}(a)|^2=\lVert s_F\rVert^2\leq\lVert x\rVert^2,\; s\in M_F\] then \[\sum_{a\in A}|\hat{x}(a)|^2\leq\lVert x\rVert^2\] holds and is called Bessel inequality. Bessel inequality shows that mapping \(\hat{x}\) maps \(H\) into \(\ell^2(A)\). The linearity of \(\hat{x}\) is obvious. If we apply Bessel inequality to \(x-y\) we see that \[\lVert \hat{x}(y)-\hat{x}(x)\rVert_2=\lVert \hat{y}-\hat{x}\rVert_2\leq\lVert y-x\rVert\] Thus \(\hat{x}\) is continuous.
    Since \[\lVert x\rVert^2=\sum_{a\in F}|\hat{x}(a)|^2,\quad F\subset A\] then \(\hat{x}\) is an isometry of \(P\) onto the dense subspace of \(\ell^2(A)\) consisting of those functions whose support is a finite set \(F\subset A\). The fact that the mapping \(x\to\hat{x}\) carries \(H\) onto \(\ell^2(A)\) is known as the Riesz-Fischer theorem, which asserts that if \(\{c_n\}\) is a sequence of complex numbers such that \[\sum_{n=-\infty}^{\infty}|c_n|^2<\infty\] then there exists an \(f\in L^2(T)\) such that \[c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)e^{-int}dt\quad(n\in Z)\]

  4. Parseval’s theorem Let \(\{u_a:a\in A\}\) be an orthonormal set in \(H\). Each of the following four conditions on \(\{u_a\}\) implies the other three:

  1. \(\{u_a\}\) is a maximal orthonormal set in \(H\). Maximal orthonormal sets are often called orthonormal bases.
  2. The set \(P\) of all finite linear combinations of members of\(\{u_a\}\) is dense in \(H\).
  3. The equality \[\sum_{a\in A}|\hat{x}(a)|^2=\lVert x\rVert^2\] holds for every \(x\in H\).
  4. The equality \[\sum_{a\in A}\hat{x}(a)\overline{\hat{y}(a)}=(x,y)\] holdsfor all \(x,y\in H\). which is known as Parseval’s identity. Since \(\hat{x}\) and \(\hat{y}\) are in \(\ell^2(A)\), hence \(\hat{x}\overline{\hat{y}}\) is in \(\ell^1(A)\), so that the sum \[\sum_{a\in A}\hat{x}(a)\overline{\hat{y}(a)}=(x,y)\] is well defined.
    If \(P\) is not dense in \(H\), then its closure \(\bar{P}\) is not all of \(H\), since if \(M \ne H\), then there exists \(y \in H\), \(y \ne 0\), such that \(y\perp M\), then \(P^{\perp}\) contains a nonzero vector. Thus \(\{u_a\}\) is not maximal when \(P\) is not dense, and (a) implies (b). If (b) holds, so does (c). Since Hilbert space identity: \[4(x,y)=\lVert x+y\rVert^2-\lVert x-y\rVert^2+i\lVert x+iy\rVert^2-i\lVert x-iy\rVert^2\] which is equally valid with \(\hat{x},\hat{y}\) in place of \(x, y\), simply because \(\ell^2(A)\) is also a Hilbert space.
    Finally, if (a) is false, there exists \(u\ne0,u\in H\) so that \((u,u_a)=0\) for all \(a\in A\). If \(x=y=u\), then \((x,y)=\lVert u\rVert^2>0\) but \(\hat{x}(a)=0,\;\; \forall a\in A\), hence (d) fails. Thus (d) implies (a). The Parseval theorem asserts that \[\sum_{n=-\infty}^{\infty}\hat{f}(n)\overline{\hat{g}(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt\] whenever \(f\in L^2(T)\) and \(g\in L^2(T)\), the series \[\sum_{n=-\infty}^{\infty}\hat{f}(n)\overline{\hat{g}(n)}\] converges absolutely; and if \(s_N\) is as in \[s_N(t)=\sum_{-N}^{N}\hat{f}(n)e^{int}\quad(N=0,1,2,\cdots)\] then \[\lim_{N\to\infty}\lVert f-s_N\rVert_2=0\] since a special case of \[\sum_{n=-\infty}^{\infty}\hat{f}(n)\overline{\hat{g}(n)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt\] yields \[\lVert f-s_N\rVert_2^2=\sum_{|n|>N}|\hat{f}(n)|^2\]
    Note that \[\lim_{N\to\infty}\lVert f-s_N\rVert_2=0\] says that every \(f\in L^2(T)\) is the \(L^2\)-limit of the partial sums of its Fourier series; i.e., the Fourier series of \(f\) converges to \(f\), in the \(L^2\)-sense.
  1. The Riesz-Fischer theorem and the Parseval theorem may be summarized by saying that the mapping \(f\to \hat{f}\) is a Hilbert space isomorphism of \(L^2(T)\) onto \(\ell^2(Z)\).

BIBLIOGRAPHY

1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.