Hilbert Space Geometry and Riesz Representation — Examples

On ${\mathbb{K}}^n$ the displayed pairing is linear in the first argument and conjugate symmetric by distributing each finite sum and applying the scalar conjugation identities of [A2], and positive definite because ${\sum }_{j<n}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2=0$ forces every $\mathrm{\mid }{x}_j\mathrm{\mid }=0$ and hence every $x_j=0$ by [A2]; the induced length is $({\sum }_{j<n}\mathrm{\mid }{x}_j{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$, a norm by [A3]. The coordinate vectors $e_j$, $j<n$, span by $x={\sum }_{j<n}{x}_j{e}_j$ and are independent by reading each coordinate, hence form an ordered basis (the empty basis if $n=0$). The finite-basis completeness theorem [A3] therefore applies; so ${\mathbb{K}}^n$ is a Hilbert space for this pairing.

On ${\mathrm{\ell }}^2(\mathbb{N};\mathbb{K})$ the pairing is the counting-measure integral of $f\bar{g}$ by [A4], so $⟨x,y⟩={\sum }_{k\ge 0}{x}_k\overline{y_k}$ with absolutely convergent series, since $2\mathrm{\mid }{x}_k\overline{y_k}\mathrm{\mid }\le \mathrm{\mid }{x}_k{\mathrm{\mid }}^2+\mathrm{\mid }{y}_k{\mathrm{\mid }}^2$ and [A4] applies to the summable right-hand side; the complex case is [A5] and the real case is the restriction of [A5] to real-valued classes, where conjugation is the identity, so in both cases the axioms of [A1] hold and the induced length is the ${\mathrm{\ell }}^2$ norm; completeness is the counting-measure instance of [A6].

On the quotient $L^2(\mu ;\mathbb{K})$ the displayed pairing is well defined on a.e. classes and satisfies the inner-product axioms with $⟨[f],[f]⟩=\mathrm{\parallel }[f]{\mathrm{\parallel }}_2^2$ by [A5] in the complex case and by the same statement restricted to real-valued classes in the real case, and completeness is [A6].

Hence ${\mathbb{K}}^n$, ${\mathrm{\ell }}^2(\mathbb{N};\mathbb{K})$ and $L^2(\mu ;\mathbb{K})$ are inner-product spaces complete for the induced norms, that is Hilbert spaces, with the pairings displayed in the statement.

The Gram matrix $G$ is invertible by [A1], so $G^{\mathsf{T}}$ is invertible and $c=(G^{\mathsf{T}}{)}^{-1}b$ is the unique solution of $G^{\mathsf{T}}c=b$; and $M$ is closed by [A2].

With $m={\sum }_j{c}_j{v}_j\in M$ one has $⟨x-m,v_i⟩=⟨x,v_i⟩-{\sum }_j{c}_j⟨v_j,v_i⟩=b_i-(G^{\mathsf{T}}c{)}_i=0$ for every $i$, and hence $⟨x-m,w⟩=0$ for every $w={\sum }_i{a}_i{v}_i\in M$ by conjugate-linearity in the second argument.

Therefore $m\in M$ and $x-m\in M^{\perp }$, so $m$ satisfies the two defining properties of the Hilbert projection and $P_{M}x=m={\sum }_j{c}_j{v}_j$.

Basis independence and uniqueness: if $w_1,\dots ,w_n$ is another independent list with the same span $M$, its Gram matrix again has nonzero determinant and the same argument gives $P_{M}x$ as a linear combination of the $w_i$ with the unique coefficient vector solving the corresponding system; since $P_{M}x\in M$ is the same vector, the two displayed formulas agree, and the coefficient vector is unique because $G^{\mathsf{T}}$ is invertible.

The integral ${\int }_{X}fd\mu$ is finite by [A2], the constant $\mu (X)$ is finite and positive, and $M$ is a closed one-dimensional subspace by [A1].

Put $c_0=\mu (X{)}^{-1}{\int }_{X}fd\mu$ and $m=c_0[\mathbf{1}]$; then $m\in M$ and $⟨[f]-m,[\mathbf{1}]⟩={\int }_X(f-c_0)d\mu ={\int }_{X}fd\mu -c_0\mu (X)=0$, so $[f]-m\in M^{\perp }$.

By the characterisation of the Hilbert projection, $P_M[f]=m=(\mu (X{)}^{-1}{\int }_{X}fd\mu )[\mathbf{1}]$; conversely, any constant $c$ with ${\int }_X(f-c)d\mu =0$ has $[f]-c[\mathbf{1}]\in M^{\perp }$, so $c=c_0$ by uniqueness of the projection, and for a probability measure the constant is the mean of $f$.

For $m\in M$ write $x-m=(x-P_{M}x)+(P_{M}x-m)$; the first summand lies in $M^{\perp }$ and the second in $M$, so the two are orthogonal and Pythagoras gives $\mathrm{\parallel }x-m{\mathrm{\parallel }}^2=\mathrm{\parallel }x-P_{M}x{\mathrm{\parallel }}^2+\mathrm{\parallel }{P}_{M}x-m{\mathrm{\parallel }}^2$.

Since $\mathrm{\parallel }{P}_{M}x-m{\mathrm{\parallel }}^2\ge 0$, step 1.1 gives $\mathrm{\parallel }x-m\mathrm{\parallel }\ge \mathrm{\parallel }x-P_{M}x\mathrm{\parallel }$ for every $m\in M$, with equality exactly at $m=P_{M}x$; hence the infimum of the distances is $\mathrm{\parallel }x-P_{M}x\mathrm{\parallel }$, attained uniquely there.

The pairing is an inner product on $c_{00}$: linearity in the first argument and conjugate symmetry are finite-sum algebra, and $⟨x,x⟩={\sum }_k\mathrm{\mid }{x}_k{\mathrm{\mid }}^2=0$ forces every coordinate $x_k$ to vanish; the induced length is the ${\mathrm{\ell }}^2$ norm of the finitely supported sequence.

Let $u^{(N)}$ be the sequence with $u_k^{(N)}=1\mathrm{/}(k+1)$ for $k<N$ and $u_k^{(N)}=0$ for $k\ge N$; each $u^{(N)}$ lies in $c_{00}$, and for $M>N$ one has $\mathrm{\parallel }{u}^{(M)}-u^{(N)}{\mathrm{\parallel }}^2={\sum }_{N\le k<M}1\mathrm{/}(k+1{)}^2={\sum }_{N<m\le M}1\mathrm{/}{m}^2$, a difference of partial sums of the convergent $p$-series, which tends to $0$ as $N,M\to \mathrm{\infty }$ by [A1]; hence $(u^{(N)})$ is Cauchy in the ${\mathrm{\ell }}^2$ norm.

Suppose $v\in c_{00}$ were a limit of $(u^{(N)})$ in the induced norm; then for each fixed $k$, Cauchy–Schwarz applied to $v-u^{(N)}$ and the $k$-th coordinate vector gives $\mathrm{\mid }{v}_k-u_k^{(N)}\mathrm{\mid }\le \mathrm{\parallel }v-u^{(N)}\mathrm{\parallel }$, so $v_k={\lim }_N{u}_k^{(N)}=1\mathrm{/}(k+1)$ for every $k$, and $v$ has infinitely many nonzero coordinates, contrary to finite support.

Hence the Cauchy sequence $(u^{(N)})$ in the inner-product space $c_{00}$ has no limit there, so $c_{00}$ is not complete for its induced norm, and the statement that every inner-product space is complete is false.

In ${\mathrm{\ell }}^p$ the function $f=e_1+e_2$ has $\mathrm{\mid }f{\mathrm{\mid }}^p=1$ at the two indices $0,1$ and $0$ elsewhere, so $\mathrm{\parallel }{e}_1+e_2{\mathrm{\parallel }}_p=(1+1{)}^{1\mathrm{/}p}=2^{1\mathrm{/}p}$, and likewise $\mathrm{\parallel }{e}_1-e_2{\mathrm{\parallel }}_p=2^{1\mathrm{/}p}$ because $\mathrm{\mid }f\mathrm{\mid }=1$ at the same two indices; in ${\mathrm{\ell }}^{\mathrm{\infty }}$ the two norms are $\mathrm{\parallel }{e}_1\pm e_2{\mathrm{\parallel }}_{\mathrm{\infty }}=1$.

The parallelogram law in ${\mathrm{\ell }}^p$ would therefore read $2\cdot 2^{2\mathrm{/}p}=4$, that is $2^{2\mathrm{/}p}=2=2^1$, which by strict monotonicity of the rational powers of base $2$ forces $2\mathrm{/}p=1$, that is $p=2$; for $p\mathrm{\ne }2$ it fails, and in ${\mathrm{\ell }}^{\mathrm{\infty }}$ the two sides are $2$ and $4$, so it fails there too.

By the Jordan–von Neumann characterisation, the norm of ${\mathrm{\ell }}^p$ with $p\mathrm{\ne }2$, and the supremum norm of ${\mathrm{\ell }}^{\mathrm{\infty }}$, are therefore not induced by any inner product on a space containing the two linearly independent coordinate vectors: the parallelogram law fails on $e_1,e_2$, and it would hold on every pair of vectors if an inducing inner product existed.

$C$ is closed and convex, and is nonempty with $0\in C$.

For $t\ge 0$ the point $c=t$ lies in $C$ with $\mathrm{\mid }t-c\mathrm{\mid }=0$, so $P_C(t)=t$; for $t<0$ and any $c\in C$ one has $\mathrm{\mid }t-c\mathrm{\mid }=c-t\ge -t=\mathrm{\mid }t-0\mathrm{\mid }$ with equality exactly at $c=0$, so $P_C(t)=0$.

Hence $P_C(t)=\max \{t,0\}$; this map is not additive, since $P_C(1)+P_C(-1)=1+0=1\mathrm{\ne }0=P_C(0)$, and it is not homogeneous either, since $P_C(-1)=0\mathrm{\ne }-1=-P_C(1)$.

Therefore the nearest-point map of a nonempty closed convex set in a Hilbert space need not be linear, so the statement refuted is false.

For $x,y\in {\mathrm{\ell }}^2$ the series in $⟨Sx,y⟩={\sum }_{n\ge 0}(Sx{)}_n\overline{y_n}={\sum }_{n\ge 1}{x}_{n-1}\overline{y_n}={\sum }_{k\ge 0}{x}_k\overline{y_{k+1}}=⟨x,Ly⟩$ converge absolutely by Cauchy–Schwarz, so $S^{\ast }=L$ by uniqueness of the adjoint.

The multiplication operator satisfies $\mathrm{\parallel }{M}_{m}f{\mathrm{\parallel }}_2\le \mathrm{\parallel }m{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }f{\mathrm{\parallel }}_2$ by [A3], and $⟨M_{m}f,g⟩=\int mf\bar{g}=\int f\overline{\stackrel{‾}{m}g}=⟨f,M_{\bar{m}}g⟩$ for all $f,g\in L^2$, so $M_m$ is bounded with $M_m^{\ast }=M_{\bar{m}}$.

For $f\in L^2(\nu )$ put $h(x):=({\int }_Y\mathrm{\mid }k(x,y){\mathrm{\mid }}^{2}d\nu (y){)}^{1\mathrm{/}2}\in [0,+\mathrm{\infty }]$. Wherever the section integral converges absolutely, the Cauchy–Schwarz inequality in the $y$-variable gives $\mid {\int }_{Y}k(x,y)f(y)d\nu (y)\mid \le {\int }_Y\mathrm{\mid }k(x,y)f(y)\mathrm{\mid }d\nu (y)\le h(x)\mathrm{\parallel }f{\mathrm{\parallel }}_2$; by Tonelli [A4] applied to the nonnegative $(\mu \times \nu )$-measurable function $\mathrm{\mid }k{\mathrm{\mid }}^2$, the function $h^2$ is $\mu$-measurable with ${\int }_{X}h(x{)}^{2}d\mu (x)=\mathrm{\parallel }k{\mathrm{\parallel }}_2^2<+\mathrm{\infty }$, so $h\in L^2(\mu )$ and $h(x)<+\mathrm{\infty }$ for $\mu$-almost every $x$. Hence $(Kf)(x)={\int }_{Y}k(x,y)f(y)d\nu (y)$ is defined and finite for $\mu$-almost every $x$, and it is $\mu$-measurable after zero extension: Tonelli [A4] makes the section integrals of the positive and negative parts of $\mathrm{Re}(kf)$ and $\mathrm{Im}(kf)$ $\mu$-measurable, and on the conull set where the integral of $kf$ is finite the real and imaginary parts of $(Kf)(x)$ are differences of these measurable functions. Since $\mathrm{\mid }Kf\mathrm{\mid }\le h\mathrm{\parallel }f{\mathrm{\parallel }}_2$ almost everywhere and $h\mathrm{\parallel }f{\mathrm{\parallel }}_2\in L^2(\mu )$, the class of $Kf$ lies in $L^2(\mu )$ with $\mathrm{\parallel }Kf{\mathrm{\parallel }}_2\le \mathrm{\parallel }k{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2$. Finally, if $k=k^{\mathrm{\prime }}$ and $f=f^{\mathrm{\prime }}$ almost everywhere, then the function $(x,y)\mapsto \mathrm{\mid }k(x,y)f(y)-k^{\mathrm{\prime }}(x,y)f^{\mathrm{\prime }}(y)\mathrm{\mid }$ is nonnegative and measurable with vanishing product integral, so for $\mu$-almost every $x$ its section vanishes $\nu$-almost everywhere by Tonelli [A4], that is, $Kf=K{f}^{\mathrm{\prime }}$ $\mu$-almost everywhere.

For $f\in L^2(\nu )$ and $g\in L^2(\mu )$ the function $(x,y)\mapsto k(x,y)f(y)\overline{g(x)}$ lies in $L^1(\mu \times \nu )$: with $h$ as in step 1.3, Tonelli and Cauchy–Schwarz in $L^2(\mu )$ give ${\int }_{X\times Y}\mathrm{\mid }k(x,y)f(y)\overline{g(x)}\mathrm{\mid }d(\mu \times \nu )={\int }_X\mathrm{\mid }g(x)\mathrm{\mid }({\int }_Y\mathrm{\mid }k(x,y)f(y)\mathrm{\mid }d\nu (y))d\mu (x)\le \mathrm{\parallel }f{\mathrm{\parallel }}_2{\int }_{X}h(x)\mathrm{\mid }g(x)\mathrm{\mid }d\mu (x)\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }h{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2=\mathrm{\parallel }k{\mathrm{\parallel }}_2\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2<+\mathrm{\infty }$. Hence Fubini [A4] applies and $⟨Kf,g⟩={\int }_X{\int }_{Y}k(x,y)f(y)\overline{g(x)}d\nu (y)d\mu (x)={\int }_{Y}f(y)\overline{{\int }_X\stackrel{‾}{k(x,y)}g(x)d\mu (x)}d\nu (y)=⟨f,K^{\ast }g⟩$ with $K^{\ast }g(y)={\int }_X\overline{k(x,y)}g(x)d\mu (x)$; applying the same estimate to the conjugate kernel $\bar{k}$, which also lies in $L^2(\mu \times \nu )$ with the same norm, gives $\mathrm{\parallel }{K}^{\ast }g{\mathrm{\parallel }}_2\le \mathrm{\parallel }k{\mathrm{\parallel }}_2\mathrm{\parallel }g{\mathrm{\parallel }}_2$, so $K^{\ast }$ is a bounded operator and is the Hilbert adjoint of $K$.

Steps 1.1, 1.2 and 2.1 exhibit the three displayed adjoints, so the claimed formulas hold, under the countable-choice hypothesis recorded in [A5].