Orthonormal Bases, Parseval and Fourier Series — Examples

The coordinate vectors are orthonormal: for all $n,m$ one has $⟨u_n,u_m⟩={\delta }_{nm}$ because each finite sum has the single surviving term $n=m$.

The space ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{F})$ is complete: if $(a^{(m)})$ is Cauchy, then for each fixed $n$ the scalars $a_n^{(m)}$ form a Cauchy sequence in the complete field $\mathbb{F}$ because $\mathrm{\mid }{a}_n^{(m)}-a_n^{(p)}\mathrm{\mid }\le \mathrm{\parallel }{a}^{(m)}-a^{(p)}{\mathrm{\parallel }}_2$, hence converge to a scalar $a_n$; by [A3] the family $a=(a_n)$ lies in ${\mathrm{\ell }}^2$ and is the limit of the sequence.

The closed linear span of the coordinate vectors is all of ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{F})$: for $a\in {\mathrm{\ell }}^2$ and $\epsilon >0$, apply [A1] with ${\epsilon }^2>0$ to obtain a finite $F$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{a}_n{\mathrm{\mid }}^2<{\epsilon }^2$. The vector $a\cdot {\mathbf{1}}_F={\sum }_{n\in F}{a}_n{u}_n$ lies in the span and satisfies $\mathrm{\parallel }a-a\cdot {\mathbf{1}}_F{\mathrm{\parallel }}_2<\epsilon$; hence every $a$ is a limit of span elements.

By steps 1.1, 1.2 and 2.1, ${\mathrm{\ell }}^2(\mathbb{N},\mathbb{F})$ is a Hilbert space and $(u_n)$ is an orthonormal family with dense span, hence a complete orthonormal family, i.e. an orthonormal basis; the expansion $a={\sum }_n{a}_n{u}_n$ is then the finite-subset expansion, and the canonical partial sums ${\sum }_{n<N}{a}_n{u}_n$ converge to $a$ because their distance to $a$ is the square root of the omitted tail ${\sum }_{n\ge N}\mathrm{\mid }{a}_n{\mathrm{\mid }}^2$, while the nondecreasing partial sums $t_N={\sum }_{n<N}\mathrm{\mid }{a}_n{\mathrm{\mid }}^2$ converge to their supremum $\mathrm{\parallel }a{\mathrm{\parallel }}_2^2$ by [A4], so the tails tend to $0$; the norm formula is the definition of $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ in [A1].

For each $r$, apply finite Gram–Schmidt to $(1,x,\dots ,x^r)$. By [A1] every residual is nonzero, and the recursive formula is independent of how far the finite list is extended, so these finite outputs are compatible and define one orthonormal sequence $(e_n{)}_{n\in \mathbb{N}}$. The successive-span identity gives $\mathrm{span}\{e_0,\dots ,e_r\}=\mathrm{span}\{1,x,\dots ,x^r\}$ for every $r$. Hence the sequence and the monomials have the same closed linear span; this is all of $L^2([-1,1])$, because the polynomials are uniformly dense in $C([-1,1])$ and the continuous functions are dense in $L^2$. Thus $(e_n)$ is a complete orthonormal family, that is, an orthonormal basis.

First element: $x_0=1$ has $\mathrm{\parallel }1{\mathrm{\parallel }}_2^2={\int }_{-1}^{1}1dx=2$, so $e_0=1\mathrm{/}\sqrt{2}$.

Second element: $v_1=x-⟨x,e_0⟩e_0$ with $⟨x,e_0⟩=\frac{1}{\sqrt{2}}{\int }_{-1}^{1}xdx=0$, so $v_1=x$ and $\mathrm{\parallel }x{\mathrm{\parallel }}_2^2={\int }_{-1}^1{x}^{2}dx=[x^3\mathrm{/}3{]}_{-1}^1=\frac{2}{3}$, giving $e_1=\sqrt{3\mathrm{/}2}x$.

Third element: $v_2=x^2-⟨x^2,e_0⟩e_0-⟨x^2,e_1⟩e_1$ with $⟨x^2,e_1⟩=\sqrt{\frac{3}{2}}{\int }_{-1}^1{x}^{3}dx=0$ and $⟨x^2,e_0⟩=\frac{1}{\sqrt{2}}{\int }_{-1}^1{x}^{2}dx=\frac{1}{\sqrt{2}}\cdot \frac{2}{3}$, so $v_2=x^2-\frac{1}{3}$; its squared norm is ${\int }_{-1}^1(x^2-\frac{1}{3}{)}^{2}dx=[x^5\mathrm{/}5-\frac{2}{9}{x}^3+\frac{1}{9}x{]}_{-1}^1=\frac{2}{5}-\frac{4}{9}+\frac{2}{9}=\frac{8}{45}$, giving $e_2=\sqrt{45\mathrm{/}8}(x^2-\frac{1}{3})=\sqrt{5\mathrm{/}8}(3x^2-1)$.

Therefore the Gram–Schmidt family of the monomials is a complete orthonormal family of $L^2([-1,1])$ beginning with $1\mathrm{/}\sqrt{2}$, $\sqrt{3\mathrm{/}2}x$ and $\sqrt{5\mathrm{/}8}(3x^2-1)$. Rescaling these three displayed unit vectors to have value $1$ at $x=1$ gives $P_0=1$, $P_1=x$, and $P_2=(3x^2-1)\mathrm{/}2$.

Orthonormality and ${\int }_0^1{h}_{j,k}=0$: each $h_{j,k}$ takes the two values $\pm 2^{j\mathrm{/}2}$ on two intervals of equal length $2^{-j-1}$, so its integral over $\mathbb{R}$ is $2^{j\mathrm{/}2}(2^{-j-1}-2^{-j-1})=0$ and $\mathrm{\parallel }{h}_{j,k}{\mathrm{\parallel }}_2^2=2^j\cdot 2\cdot 2^{-j-1}=1$. Two distinct such functions either have disjoint supports, giving pairing $0$, or have nested supports $I_{j^{\mathrm{\prime }},k^{\mathrm{\prime }}}\subseteq I_{j,k}$ with $j^{\mathrm{\prime }}>j$, and the finer interval lies wholly in one half of the coarser interval, so the coarser function is constant there. Then the pairing is $\pm 2^{j\mathrm{/}2}\int h_{j^{\mathrm{\prime }},k^{\mathrm{\prime }}}=0$ because the integral of the finer function vanishes; and $⟨1,h_{j,k}⟩=\int h_{j,k}=0$, while $\mathrm{\parallel }1{\mathrm{\parallel }}_2^2=1$.

Every continuous $f:[0,1]\to \mathbb{C}$ is uniformly approximated on $[0,1)$ by functions constant on the level-$J$ dyadic intervals: by uniform continuity choose $\delta >0$ with $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }x-y\mathrm{\mid }<\delta$, choose $J$ with $2^{-J}<\delta$, and let $s_J$ be the function that on each level-$J$ interval takes the value of $f$ at its left endpoint; then ${\sup }_{x\in [0,1)}\mathrm{\mid }f(x)-s_J(x)\mathrm{\mid }\le \epsilon$ and hence $\mathrm{\parallel }f-s_J{\mathrm{\parallel }}_{L^2((0,1))}\le \epsilon$.

For every $J\ge 0$ the linear span of $\{1\}\cup \{h_{j,k}:j<J\}$ is exactly the space $V_J$ of functions constant on each level-$J$ dyadic interval: the two functions $1_{[a,m)}$ and $1_{[m,b)}$ of a level-$(j+1)$ interval inside a level-$j$ interval are $(1_{[a,b)}\pm 2^{-j\mathrm{/}2}{h}_{j,k})\mathrm{/}2$, so by induction every level-$J$ dyadic indicator lies in the span, and conversely every $h_{j,k}$ with $j<J$ is a linear combination of level-$J$ indicators; hence the span is contained in $V_J$ and contains all its indicators.

Therefore the closed linear span of the family is $L^2((0,1))$: it contains $V_J$ for every $J$ by step 2.1, hence by step 1.2 it contains the restrictions of $C([0,1])$, and their $L^2((0,1))$-closure is $L^2((0,1))$.

Since the family is orthonormal by step 1.1 and its closed linear span is all of $L^2((0,1))$ by step 3.1, it is an orthonormal basis of $L^2((0,1))$ in both the real and the complex case.

The zeroth coefficient vanishes: $\widehat{f}(0)={\int }_0^{1}f(t)dt={\int }_0^{1\mathrm{/}2}tdt+{\int }_{1\mathrm{/}2}^1(t-1)dt=\frac{1}{8}+[\frac{t^2}{2}-t{]}_{1\mathrm{/}2}^1=\frac{1}{8}-\frac{1}{2}+\frac{3}{8}=0$.

For $k\mathrm{\ne }0$, write $e^{-2\pi ikt}=\cos (2\pi kt)-i\sin (2\pi kt)$ and integrate by parts on the two halves. Integrating $t\cos (2\pi kt)$ and $(t-1)\cos (2\pi kt)$ with antiderivatives $\frac{\sin (2\pi kt)}{2\pi k}$ and $\frac{\cos (2\pi kt)}{4{\pi }^2{k}^2}$ gives ${\int }_0^{1\mathrm{/}2}t\cos (2\pi kt)dt=\frac{\sin (\pi k)}{4\pi k}+\frac{\cos (\pi k)-1}{4{\pi }^2{k}^2},{\int }_{1\mathrm{/}2}^1(t-1)\cos (2\pi kt)dt=\frac{\sin (\pi k)}{4\pi k}+\frac{1-\cos (\pi k)}{4{\pi }^2{k}^2},$ so ${\int }_0^{1}f(t)\cos (2\pi kt)dt=\frac{\sin (\pi k)}{2\pi k}=0$; and integrating $t\sin (2\pi kt)$ and $(t-1)\sin (2\pi kt)$ with antiderivatives $-\frac{t\cos (2\pi kt)}{2\pi k}+\frac{\sin (2\pi kt)}{4{\pi }^2{k}^2}$ and $-\frac{(t-1)\cos (2\pi kt)}{2\pi k}+\frac{\sin (2\pi kt)}{4{\pi }^2{k}^2}$ gives ${\int }_0^{1\mathrm{/}2}t\sin (2\pi kt)dt=-\frac{\cos (\pi k)}{4\pi k}+\frac{\sin (\pi k)}{4{\pi }^2{k}^2},{\int }_{1\mathrm{/}2}^1(t-1)\sin (2\pi kt)dt=-\frac{\cos (\pi k)}{4\pi k},$ so ${\int }_0^{1}f(t)\sin (2\pi kt)dt=-\frac{\cos (\pi k)}{2\pi k}=-\frac{(-1{)}^k}{2\pi k}$.

Adding the real and imaginary contributions, $\widehat{f}(k)=-i{\int }_0^{1}f(t)\sin (2\pi kt)dt=\frac{i(-1{)}^k}{2\pi k}=\frac{(-1{)}^{k+1}}{2\pi ik}$ for every $k\mathrm{\ne }0$.

Parseval's identity gives $\frac{1}{12}=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2={\sum }_{k\mathrm{\ne }0}\frac{1}{4{\pi }^2{k}^2}=\frac{1}{2{\pi }^2}{\sum }_{k\ge 1}\frac{1}{k^2}$, where $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2={\int }_0^{1}f(t{)}^{2}dt=2{\int }_0^{1\mathrm{/}2}{t}^{2}dt=\frac{1}{12}$; multiplying by $2{\pi }^2$ gives ${\sum }_{k\ge 1}{k}^{-2}={\pi }^2\mathrm{/}6$.

The coefficients of steps 1.1 and 2.1 are the displayed ones, the Fourier series converges to $f$ in mean square, and the Parseval computation of step 3.1 yields the Basel sum; the endpoint values of the representative are irrelevant, and the claim is an $L^2$ statement only.

The mean vanishes: $\widehat{s}(0)={\int }_0^{1}s(t)dt=\frac{1}{2}-\frac{1}{2}=0$.

For $k\mathrm{\ne }0$ write $e^{-2\pi ikt}=\cos (2\pi kt)-i\sin (2\pi kt)$. The cosine part vanishes: ${\int }_0^{1\mathrm{/}2}\cos (2\pi kt)dt=\frac{\sin (\pi k)}{2\pi k}=0$ and ${\int }_{1\mathrm{/}2}^1\cos (2\pi kt)dt=\frac{\sin (2\pi k)-\sin (\pi k)}{2\pi k}=0$, and the signs $\pm 1$ of $s$ multiply these to give ${\int }_0^{1}s(t)\cos (2\pi kt)dt=0$ as well. For the sine part, the antiderivative $-\frac{\cos (2\pi kt)}{2\pi k}$ gives ${\int }_0^{1}s(t)\sin (2\pi kt)dt=\frac{1-(-1{)}^k}{2\pi k}-\frac{(-1{)}^k-1}{2\pi k}=\frac{1-(-1{)}^k}{\pi k}\mathrm{.}$

Therefore $\widehat{s}(k)=-i{\int }_0^{1}s(t)\sin (2\pi kt)dt=-\frac{i(1-(-1{)}^k)}{\pi k}=\frac{1-(-1{)}^k}{\pi ik}$ for every $k\mathrm{\ne }0$; the coefficient vanishes exactly for even $k$ and is nonzero for odd $k$.

Parseval's identity gives $1=\mathrm{\parallel }s{\mathrm{\parallel }}_2^2={\sum }_{k\in \mathbb{Z}}\mathrm{\mid }\widehat{s}(k){\mathrm{\mid }}^2$, because $\mathrm{\mid }s\mathrm{\mid }=1$ and the domain has measure one; and $\mathrm{\mid }\widehat{s}(k){\mathrm{\mid }}^2=\frac{(1-(-1{)}^k{)}^2}{{\pi }^2{k}^2}$ equals $\frac{4}{{\pi }^2{k}^2}$ for odd $k$ and $0$ for even $k$. Hence $1=\frac{4}{{\pi }^2}{\sum }_{k\text{ odd}}\frac{1}{k^2}=\frac{8}{{\pi }^2}{\sum }_{k\text{ odd},k\ge 1}\frac{1}{k^2}$, and multiplying by ${\pi }^2\mathrm{/}8$ gives ${\sum }_{k\text{ odd},k\ge 1}{k}^{-2}={\pi }^2\mathrm{/}8$.

The coefficients of steps 1.1 and 2.1 are the displayed ones, [A6] gives convergence of the symmetric Fourier partial sums to $s$ in mean square, and step 3.1 evaluates the associated square sum as ${\pi }^2\mathrm{/}8$; all statements are about the $L^2$ class, and no pointwise or endpoint convergence is asserted.