Recall that a Banach space is a normed space that is complete in its norm, and a Hilbert space is an inner product space that is complete in the norm \(\norm x\coloneqq\inner xx^{1/2}\); every Hilbert space is a Banach space. In a Hilbert space \(H\), closed subspaces have orthogonal complements and orthogonal projections, and for an orthonormal set \(\theset{e_n}\) Bessel’s inequality \(\sum_n \abs{\inner{x}{e_n}}^2\leq\norm x^2\) holds for every \(x\in H\). Parseval’s identity, equality for every \(x\), holds exactly when \(\theset{e_n}\) is an orthonormal basis. The Riesz–Fischer theorem gives the converse: for an orthonormal set \(\theset{e_n}\) and a square-summable sequence \((c_n)\), \(\sum_n c_ne_n\) converges in \(H\) to an element with coefficients \(c_n\). The Riesz representation theorem states that every bounded linear functional on \(H\) is \(x\mapsto\inner{x}{y}\) for a unique \(y\in H\), with the inner product linear in its first argument.
$L^p$ spaces are Banach spaces (Tag PR-L35O7)
Let \((X,\mcm,\mu)\) be a measure space and let \(1\leq p\leq\infty\). Then \(L^p(X,\mu)\), whose elements are classes of functions modulo equality \(\mu\)-almost everywhere, with the norm \(\norm{\cdot}_p\), is a Banach space.
First let \(1\leq p<\infty\), and let \((f_n)\) be Cauchy in \(L^p\). Choose a subsequence \((f_{n_k})\) such that \[ \norm{f_{n_{k+1}}-f_{n_k}}_p\leq 2^{-k} \qquad(k\geq1). \] Set \[ g_k\coloneqq\abs{f_{n_{k+1}}-f_{n_k}}, \qquad S_N\coloneqq\sum_{k=1}^N g_k. \] Minkowski’s inequality gives \[ \norm{S_N}_p\leq\sum_{k=1}^N2^{-k}\leq1. \] Since \(S_N\uparrow S\coloneqq\sum_{k\geq1}g_k\), the monotone convergence theorem yields \[ \int_X S^p\dmu =\lim_{N\to\infty}\int_X S_N^p\dmu \leq1. \] Thus \(S<\infty\) almost everywhere. Outside a null set, the series \[ \sum_{k\geq1}\big(f_{n_{k+1}}(x)-f_{n_k}(x)\big) \] converges absolutely; there, define \(f(x)\) so that this sum equals \(f(x)-f_{n_1}(x)\), and put \(f\coloneqq0\) on the exceptional null set.
For every \(k\) and \(N>k\), \[ \abs{f_{n_N}-f_{n_k}} \leq\sum_{j=k}^{N-1}g_j. \] Letting \(N\to\infty\) and applying Fatou’s lemma together with Minkowski’s inequality gives \[ \norm{f-f_{n_k}}_p \leq\sum_{j=k}^{\infty}\norm{g_j}_p \leq\sum_{j=k}^{\infty}2^{-j} =2^{1-k}. \] Hence \(f_{n_k}\to f\) in \(L^p\). Given \(\varepsilon>0\), choose \(k\) with \(\norm{f_{n_k}-f}_p<\varepsilon/2\) and \(n_k\geq N\), where \(N\) is such that \(\norm{f_n-f_m}_p<\varepsilon/2\) for all \(m,n\geq N\). Then \(\norm{f_n-f}_p<\varepsilon\) for all \(n\geq N\), so \(f_n\to f\) in \(L^p\).
Now let \(p=\infty\), and again choose a subsequence with \[ \norm{f_{n_{k+1}}-f_{n_k}}_\infty\leq2^{-k}. \] For each \(k\) there is a null set \(Z_k\) outside which \[ \abs{f_{n_{k+1}}-f_{n_k}}\leq2^{-k}. \] Outside the null set \(Z\coloneqq\bigcup_kZ_k\), the series of successive differences converges uniformly by the Weierstrass \(M\)-test. Hence \((f_{n_k})\) converges uniformly on \(X\setminus Z\) to a bounded measurable function \(f\), extended by \(0\) on \(Z\), and \[ \norm{f-f_{n_k}}_\infty\leq2^{1-k}. \] As in the case \(p<\infty\), since \((f_n)\) is Cauchy, \(f_n\to f\) in \(L^\infty\).
Duals of $L^p$ spaces (Tag T-5BFVS)
Let \((X,\mcm,\mu)\) be a measure space, let \(1\leq p<\infty\), and let \(q\in(1,\infty]\) satisfy \(\frac1p+\frac1q=1\). If \(p=1\), assume that \(\mu\) is $\sigma$-finite. Then the map \[ L^q(X,\mu)\to \dualof{L^p(X,\mu)},\qquad g\mapsto\qty{f\mapsto\int_X fg\dmu}, \] is an isometric isomorphism onto the space \(\dualof{L^p(X,\mu)}\) of continuous linear functionals on \(L^p(X,\mu)\) with the dual norm [@Fol13].
Inner products are jointly continuous (Tag PR-KTZZ5)
Let \(H\) be an inner product space with inner product \(\inner{\cdot}{\cdot}\) and norm \(\norm{x}\coloneqq\inner{x}{x}^{1/2}\). If \(x_k\to x\) and \(y_k\to y\) in \(H\), then \(\inner{x_k}{y_k} \to \inner{x}{y}\).
Add and subtract \(\inner{x}{y_k}\): \[ \abs{\inner{x_k}{y_k} - \inner{x}{y} } = \abs{\inner{x_k - x}{y_k} + \inner{x}{y_k - y} } \leq \norm{x_k - x}\norm{y_k} + \norm{x}\norm{y_k - y} \] by the triangle inequality and the Cauchy–Schwarz inequality. Since \(y_k \to y\), the sequence \((\norm{y_k})\) is bounded, and \(\norm{x_k - x} \to 0\), so the first term tends to \(0\). Since \(\norm{y_k - y}\to 0\), the second term tends to \(0\).
Parseval's theorem for orthonormal families (Tag T-5AALA)
Let \(\mch\) be a Hilbert space and let \((u_n)_{n\in A}\) be a family in \(\mch\), indexed by a set \(A\), with \(\inner{u_n}{u_m}=0\) for \(n\neq m\) and \(\norm{u_n}=1\) for all \(n\). For each \(x\in\mch\) only countably many \(\inner{x}{u_n}\) are nonzero (T-4BDE3), and sums over \(A\) below are taken over those indices. The following are equivalent:
(Completeness) If \(x\in\mch\) and \(\inner{x}{u_n}=0\) for all \(n\in A\), then \(x=0\).
(Parseval’s identity) For every \(x\in\mch\), \[ \sum_{n\in A} \abs{ \inner{x}{u_n} }^2 = \norm{x}^2 . \]
(Expansion) For every \(x\in\mch\), \[ x = \sum_{n\in A} \inner{x}{u_n}u_n , \] with the series converging in norm in any enumeration of its nonzero terms; and if \(x=\sum_{n\in A}c_nu_n\) for scalars \(c_n\), only countably many nonzero, then \(c_n=\inner{x}{u_n}\) for all \(n\).
Parseval's identity (Tag FT-NFMJW)
Let \(H\) be a Hilbert space and let \((e_k)_{k\geq1}\) be an orthonormal sequence in \(H\). For every \(x\in H\), Bessel’s inequality holds: \[ \sum_{k\geq1} \abs{\inner{x}{e_k}}^2 \le \norm{x}^2 . \] If \((e_k)_{k\geq1}\) is a basis of \(H\), then equality holds for every \(x\in H\): \[ \sum_{k\geq1} \abs{\inner{x}{e_k}}^2 = \norm{x}^2 . \]
Riesz representation theorem for Hilbert spaces (Tag T-LDCZB)
Let \(H\) be a Hilbert space and let \(\Lambda\in \dualof{H}\) be a continuous linear functional on \(H\). Then there exists a unique \(y \in H\) such that \[ \Lambda(x) = \inner{x}{y} \quad\text{for all } x\in H . \]
Riesz representation theorem for Hilbert spaces (Tag FF-UT5GL)
Let \(H\) be a Hilbert space over \(\CC\) with inner product \(\inner{\cdot}{\cdot}\) linear in the first argument, and let \(\varphi \in \dualof{H}\) be an element of its dual. Then there exists a unique \(f\in H\) such that \(\varphi(x) = \inner{x}{f}\) for all \(x\in H\), and \(\norm{f}_H = \norm{\varphi}_{\dualof{H}}\), the dual norm of \(\varphi\).
Riesz--Fischer theorem (Tag T-J3AN3)
Let \(\mathcal H\) be a Hilbert space, let \(U = (u_n)_{n\geq1}\) be an orthonormal sequence in \(\mathcal H\), and let \(\mathcal H_U\subseteq\mathcal H\) be the closure of the linear span of \(\theset{u_n\suchthat n\geq1}\). Then the coefficient map \[ \begin{aligned} \mathcal H_U &\longrightarrow \ell^2(\NN) \\ x &\longmapsto (\inner{x}{u_n})_{n\geq1} \end{aligned} \] is an isometric isomorphism. Equivalently, for every \((a_n)_{n\geq1}\in\ell^2(\NN)\) there is a unique \(x\in\mathcal H_U\) such that \[ a_n=\inner{x}{u_n}\quad\text{for all }n, \qquad\text{and then}\qquad \norm{x}^2=\sum_{n=1}^\infty\abs{a_n}^2. \]
In particular, if \(U\) is complete, so that \(\mathcal H_U=\mathcal H\), then the coefficient map \(\mathcal H\to\ell^2(\NN)\) is an isometric isomorphism.
For \(a=(a_n)\in\ell^2(\NN)\), set \[ x_N\coloneqq\sum_{n=1}^N a_nu_n. \] Orthonormality gives, for \(M<N\), \[ \norm{x_N-x_M}^2=\sum_{n=M+1}^N\abs{a_n}^2, \] so \((x_N)\) is Cauchy and converges, by completeness of \(\mathcal H\), to some \(x\), which lies in \(\mathcal H_U\) because each \(x_N\) lies in the span of \(U\). Continuity of the inner product yields \(\inner{x}{u_n}=a_n\) for every \(n\), and taking limits in \(\norm{x_N}^2=\sum_{n=1}^N\abs{a_n}^2\) gives \[ \norm{x}^2=\sum_{n=1}^\infty\abs{a_n}^2. \]
Conversely, let \(x\in\mathcal H_U\) and put \(a_n\coloneqq\inner{x}{u_n}\). Bessel’s inequality T-4BDE3 gives \(a\in\ell^2(\NN)\). Applying the construction above produces \(y\in\mathcal H_U\) with the same coefficients. Then \(x-y\in\mathcal H_U\) is orthogonal to every \(u_n\), hence to the span of \(U\) and therefore to its closure \(\mathcal H_U\); in particular \(x-y\) is orthogonal to itself, so \(x=y\). This proves bijectivity and the norm identity.
(Parallelogram law)
A norm on a vector space is induced by an inner product if and only if it satisfies \(\norm{x+y}^2+\norm{x-y}^2 = 2\norm x^2+2\norm y^2\) for all \(x, y\).
($L^p$ is a Hilbert space only for $p=2$)
Let \(E,F\) be disjoint measurable sets with \(0<\mu(E),\mu(F)<\infty\), and let \(f\coloneqq\chi_E/\norm{\chi_E}_p\) and \(g\coloneqq\chi_F/\norm{\chi_F}_p\). For \(1\leq p<\infty\), \(\norm{f\pm g}_p = 2^{1/p}\), so the parallelogram law reads \(2\cdot2^{2/p} = 4\), which holds only for \(p=2\). For \(p=\infty\), \(\norm{f\pm g}_\infty=1\) and the law reads \(2=4\). So \(L^p(\mu)\) is not a Hilbert space for \(p\neq2\) on such a measure space.