There is no translation-invariant measure on an infinite-dimensional space

One difficulty in carrying out analysis on an infinite-dimensional normed vector space, e.g. $L^p(\mathbb{R})$, is that there is no analogue of Lebesgue measure on such a space. This comes from the observation that an open ball in an infinite-dimensional normed vector space contains infinitely many disjoint open balls of the same radius. We first give a formal statement of our goal.

Theorem. Let $X$ be an infinite-dimensional normed vector space. Then there is no translational-invariant Borel measure $\mu$ on $X$ such that $\mu(U) > 0$ for every nonempty open set $U$ and $\mu(U_0) < \infty$ for some open set $U_0$.

To prove the theorem, we recall Riesz's Lemma.

Lemma (Riesz). Let $X$ be a normed vector space and $Y$ be a proper closed subspace. For every $\epsilon > 0$, there exists $x \in X$ such that $\|x\| = 1$ and $\mathrm{dist}(x, Y) = \inf\{\|x - y\|: y \in Y\} \geq 1 - \epsilon$.

Proof of Theorem: Suppose $\mu$ is a translation-invariant Borel measure on $X$ with $\mu(U) > 0$ for every nonempty open $U \subseteq X$. Fix a nonempty open set $U$ in $X$. Then there exists $r > 0$ and $x_0 \in X$ such that $W = r(U + x_0)$ contains $B = \{x \in X: \|x\| < 2 \}$. Using Riesz's Lemma and induction, we can find a sequence $\{x_n\}$ of unit vectors in $X$ such that $\mathrm{dist}(x_{n + 1}, \mathrm{span}\{x_1, \ldots, x_n\}) \geq 1/2$ for all $n \geq 1$. Hence $\{x_n\} \subseteq B$ and $\|x_n - x_m\| \geq 1/2$ whenever $n \neq m$. Take $B_n = \{x \in X: \|x - x_n\| < 1/2 \}$, then $\{B_n\}$ is a disjoint collection of open balls contained in $B$ of radius half. Now let $U_n = \frac{1}{r}B_n - x_0$, then $\{U_n\}$ is a disjoint collection of open balls contained in $U$ of the same radius. It follows that \[ \mu(U) \geq \mu(\bigcup_{n = 1}^\infty U_n) = \sum_{n = 1}^\infty \mu(U_n) = \sum_{n = 1}^\infty \mu(U_1) \] by translation-invariance of $\mu$. But $\mu(U_1) > 0$, forcing $\mu(U) = \infty$.

Reference:
Fabian, Habala, Hajek, Montesinos, Zizler - Banach Space Theory: The Basis for Linear and Nonlinear Analysis

A criterion for writing a functional as a linear combination

Let $X$ be a vector space over $\mathbb{F}$ with dual space $X'$ and fix $\phi_1, \ldots, \phi_n \in X'$. We have a useful criterion for when a linear functional $f \in X'$ lies in the span of $\phi_1, \ldots, \phi_n$.

Theorem 1. Let $f \in X'$. Then $f \in \mathrm{span}\{\phi_1, \ldots, \phi_n\}$ if and only if $\cap_{i = 1}^n \ker(\phi_i) \subseteq \ker(f)$.

To prove this, we first establish a general result on when a linear map can factor through another linear map.

Theorem 2. Suppose we are given linear maps $T: X \rightarrow Y$ and $S: X \rightarrow Z$. Then there exists a linear map $R: Z \rightarrow Y$ such that $T = R \circ S$ if and only if $\ker(S) \subseteq \ker(T)$.

Proof: (=>) Clear. (<=) Suppose $\ker(S) \subseteq \ker(T)$. Define $R': \mathrm{im}(S) \rightarrow Y$ by $R'(S(x)) = T(x)$. If $S(x) = S(y)$, then $x - y \in \ker(S)$, so $x - y \in \ker(T)$, i.e. $T(x) = T(y)$. Hence $R'$ is well-defined. It is easy to check that $R'$ is linear. Now let $P$ be the projection of $Z$ onto $\mathrm{im}(S)$ and take $R = R' \circ P$. Then for all $x \in X$, $R \circ S(x) = R'(P(S(x))) = R'(S(x)) = T(x)$. 

Proof of Theorem 1: (=>) Clear. (<=) Take $T: X \rightarrow \mathbb{F}$ and $S: X \rightarrow \mathbb{F}^n$ defined by $T(x) = f(x)$ and $S(x) = (\phi_1(x), \ldots, \phi_n(x))$. Then $\ker(S) = \cap_{i = 1}^n \ker(\phi_i) \subseteq \ker(f) = \ker(T)$. By Theorem 2 there exists $R: \mathbb{F}^n \rightarrow \mathbb{F}$ such that $T = R \circ S$ or $f(x) = R(\phi_1(x), \ldots, \phi_n(x))$ for all $x \in X$. Writing $R(y_1, \cdots, y_n) = c_1 y_1 + \cdots + c_n y_n$, we see that $f = c_1 \phi_1 + \cdots + c_n \phi_n$.

Reference:
Holmes - Geometric Functional Analysis and its Applications

Weak* separation theorem and Goldstine's theorem

Recall a version of separation theorem for a Banach space is the following.

Theorem 1. Let $X$ be a Banach space. Let $C$ be a nonempty open convex subset of $X$ and $x_0 \in X \backslash C$. Then there exists $\ell_0 \in X^*$ such that $\sup_{x \in C} \Re \ell_0(x) < \Re \ell_0(x_0)$.

There is a separation theorem in the weak* topology setting.

Theorem 2. Let $X$ be a Banach space. Let $C$ be a nonempty weak*-closed convex subset of $X^*$ and $\ell_0 \in X^* \backslash C$. Then there exists $x_0 \in X$ such that $\sup_{\ell \in C} \Re \ell(x_0) < \Re \ell_0(x_0)$.
Proof: Since $X^* \backslash C$ is weak*-open, we can find $x_1, \ldots, x_n \in X$ and $\epsilon > 0$ such that $U = \{\ell \in X^* : |\ell(x_i)| < \epsilon, i = 1, \ldots, n\}$ satisfies $\ell_0 + U \subset X^* \backslash C$. This implies that $\ell_0$ does not lie in the convex open set $C + U$. By Theorem 1 there exists $\psi \in X^{**}$ such that $\Re \psi(\ell_0) > \sup_{\ell \in C + U} \Re \psi(\ell) \geq \sup_{\ell \in C} \Re \psi(\ell)$. We want to show that $\cap_{i = 1}^n \ker(x_i) \subseteq \ker(\psi)$, since then $\psi$ is a linear combination of $x_1, \ldots, x_n$, thus $x_0 = \psi \in X$ and we are done. Let $\ell \in X^*$ with $\ell(x_i)$ for all $1 \leq i \leq n$. Note that $t\ell \in U$ for all $t \in \mathbb{R}$. Fix any $c \in C$. Then $M := \sup_{f \in U} \psi(f) < \psi(\ell_0) - \psi(c)$ is finite. Now for all $t \in \mathbb{R}$, $\psi(t\ell) \leq M$, i.e. $\psi(\ell) \leq M/t$ for all $t > 0$ and $\psi(\ell) \geq M/t$ for all $t < 0$. Therefore $\psi(\ell) = 0$.

The following is an application of the weak* separation theorem. (Notation: $B_Y = \{y \in Y: \|y\| \leq 1\}$.)

Theorem (Goldstine). The weak* closure of $B_X$ in $X^{**}$ is $B_{X^{**}}$.
Proof: First, $B_X \subseteq B_{X^{**}}$ and $B_{X^{**}}$ is weak*-compact by Alaoglu's Theorem. Hence the weak* closure $\tilde{B}$ of $B_X$ is a subset of $B_{X^{**}}$. Suppose the inclusion is proper and choose $w \in B_{X^{**}} \backslash \tilde{B}$. Apply Theorem 2 to find an $\ell \in X^*$ and $\epsilon > 0$ such that $\sup_{\psi \in \tilde{B}} \Re \psi(\ell_0) \leq \Re w(\ell_0) - \epsilon$. In particular, $\Re \ell_0(x) \leq \Re w(\ell_0) - \epsilon$ for all $x \in B_X$. This implies that $|\ell_0(x)| \leq \Re w(\ell_0) - \epsilon$ for all $x \in B_X$. Replace $w$ by some $e^{-i\theta}w$, we have $|\ell_0(x)| \leq |w(\ell_0)| - \epsilon$ for all $x \in B_X$. But then $\|\ell_0\| \leq |w(\ell_0)| - \epsilon \leq \|w\|\|\ell_0\| - \epsilon = \|\ell_0\| - \epsilon$. This is a contradiction.

Reference:
Fabian, Habala, Hajek,  Santalucia, Pelant, Zizler - Functional Analysis and Infinite Dimensional Geometry

Weak*-closed subspaces as dual spaces

Let $X$ be a Banach space and $M$ be a subspace of $X^*$ closed under the weak* topology. We show that M can be identified as the dual space of some Banach space.

Let $M_\perp = \{x \in X: f(x) = 0 \,\, \forall f \in M\} = \bigcap_{f \in M} \ker(f)$. This is a (weakly) closed subspace of $X$. Then we can form the quotient Banach space $X/M_\perp$ with the norm $\|x + M_\perp\| = \inf_{m \in M_\perp} \|x - m\|$.

Theorem. $M$ is isometrically isomorphic to $(X/M_\perp)^*$.
Proof: Define a map $\Phi: M \rightarrow (X/M_\perp)^*$ by $[\Phi(f)](x + M_\perp) = f(x)$ for all $f \in M$. Firstly, $\Phi(f)$ is a well-defined linear functional on $X/M_\perp$ because $f$ vanishes on $M_\perp$ whenever $f \in M$. Secondly, $\Phi(f)$ is bounded as \[ |\Phi(f)(x + M_\perp)| = |f(x)| = |f(x - m)| \leq \|f\|\|x - m\| \,\, \forall m \in M_\perp, \] so $|\Phi(f)(x + M_\perp)| \leq \|f\|\|x + M_\perp\|$. Hence $\Phi$ is well-defined.
Clearly $\Phi$ is linear. From the above we have $\|\Phi(f)\| \leq \|f\|$. Let $\epsilon > 0$. Choose some $x \in X$ with $\|x\| \leq 1$ and $|f(x)| > \|f\| - \epsilon$. Then $\|x + M_\perp\| \leq 1$ and $|[\Phi(f)](x + M_\perp)| = |f(x)|  > \|f\| - \epsilon$. Thus $\|\Phi(f)\| = \|f\|$. It follows that $\Phi$ is an isometry.
It remains to show that $\Phi$ is surjective. Let $\ell \in (X/M_\perp)^*$ be given. Define $f \in X^*$ by $f(x) = \ell(x + M_\perp)$, i.e. $f = \ell \circ \pi$ where $\pi$ is the quotient map. Then $f$ vanishes on $M_\perp$ by definition, hence $f \in M$ by the following lemma and it is clear that $\Phi(f) = \ell$.

Lemma. If $f \in X^*$ vanishes on $M_\perp$, then $f \in M$.
Proof: Suppose $f \in X^* \backslash M$. By Hahn-Banach Theorem and weak*-closedness we can find some $x \in X$ such that $f(x) = 1$ and $g(x) = 0$ whenever $g \in M$. But then $x \in M_\perp$ and so $f(x) = 0$, contradiction.

This result implies the following important observation in the theory of operator algebras.

Theorem. Every von Neumann algebra is the dual space of some Banach space.
Proof: Let $M$ be a von Neumann algebra on a Hilbert space $\mathcal{H}$. Then $M$ is a $\sigma$-weakly closed subalgebra of $B(\mathcal{H})$, i.e. it is a weak*-closed subalgebra of $B(\mathcal{H})$ where the weak* topology of $B(\mathcal{H})$ comes from the fact that the dual space of the trace-class operators $L^1(\mathcal{H})$ on $\mathcal{H}$ can be identified with $B(\mathcal{H})$. By our main theorem, $M \cong (L^1(\mathcal{H})/M_\perp)^*$.

A predual of a von Neumann algebra $M$ is a Banach space $X$ such that $M = X^*$. The above result says that every von Neumann algebra has a predual. In fact, the predual is unique by a result of Sakai. Furthermore, the property that having a predual characterizes von Neumann algebras among the class of C*-algebras.

Theorem (Sakai).  Suppose $A$ is a W*-algebra, i.e. $A$ is a C*-algebra and the dual space of some Banach space. Then $A$ is *-isomorphic to a von Neumann algebra over some Hilbert space.

Open mapping theorem in terms of norm estimates

Here we present a simple but useful corollary of the Open Mapping Theorem in functional analysis.

Notation: $B_X(a, r) = \{x \in X: \|x - a\| < r\}$.

Proposition. Suppose $X$ and $Y$ are Banach spaces. Let $f: X \rightarrow Y$ be a surjective bounded linear map. Then there exists some $\delta > 0$ such that $\|f(x)\| \geq \delta\|x\|$ for all $x \in X$.

Proof: By the Open Mapping Theorem, $f$ is an open map. Hence we can find some $\delta > 0$ such that $B_Y(0, \delta) \subseteq f(B_X(0, 1))$, so $f(X \backslash B(0, 1)) \subseteq Y \backslash B(0, \delta)$, i.e. $\|f(x)\| \geq \delta$ whenever $\|x\| \geq 1$. Therefore $\|f(x)\| = \|x\|\|f(x/\|x\|)\| \geq \delta\|x\|$.

Here is an application of the above result in the theory of Fourier series.

Theorem. The Fourier transform $\mathcal{F}: L^1(\mathbb{T}) \rightarrow c_0(\mathbb{Z})$ is not surjective.

Proof: Suppose not. By the Proposition, there exists $\delta > 0$ such that $\|\hat{f}\|_\infty \geq \delta\|f\|_1$. But if $\{D_N\}_{N = 0}^\infty$ is the Dirichlet kernel, i.e.

\[ D_N(x) = \sum_{n = -N}^N e^{inx} \]

then we have $\|D_N\|_1 \rightarrow \infty$ and $\|\widehat{D_N}\|_\infty \geq \delta\|D_N\|_1 \rightarrow \infty$, which is absurd.

Boundedness of weakly convergent sequences

Let X be a Banach space. A net $\{x_\alpha\}$ in X is said to weakly converge to $x$ if $\ell(x_\alpha) \rightarrow \ell(x)$ for all $\ell \in X^*$. Clearly, every norm convergent net is weakly convergent (with the same limit). We know that a norm convergent sequence is necessarily bounded. It is natural to ask boundedness still hold for weakly convergent sequences.

Theorem. Every weakly convergent sequence in X is bounded.
Proof: Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$, since the sequence $\{\ell(x_n)\}$ is convergent, $\{T_n(\ell)\}$ is a bounded set. By Uniform Boundedness Principle,
\[ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\| < \infty, \] i.e. $\{x_n\}$ is bounded.

However, the above result is no longer true when sequences are replaced by nets. To give a counterexample, let $X = L^1(\mathbb{R})$. Denote the characteristic function of a subset A by $\chi_A$. Let $\mathcal{O}$ be the collection of all open neighborhoods of 0 in $\mathbb{R}$ of finite Lebesgue measure with the ordering given by $U \prec V$ iff $U \supseteq V$. Consider the net $\{\chi_U: U \in \mathcal{O}\}$ in X. It is not bounded since $\|\chi_{(-n, n)}\|_1 = 2n \rightarrow \infty$ as $n \rightarrow \infty$. On the other hand, it converges weakly to 0. Fix any $f \in X^* = L^\infty(\mathbb{R})$. Choose M > 0 such that $|f| \leq M$ almost everywhere. For every $\epsilon > 0$, we can choose some $V \in \mathcal{O}$ with $m(V) < \epsilon/M$, then $|\int f\chi_U dm| \leq M m(U) \leq M m(V) < \epsilon$ whenever $U \succ V$.