Linear functionals on the space of measurable functions.

Question

Let us consider the following space of all measurable functions on $[0,1]$ with the metric: $$d(f, g) = \int_{0}^{1} \frac{|f-g|}{1 + |f-g|}d\mu.$$ It is the metric of convergence in measure on $[0,1]$ ($\mu$ - standard Lebesgue measure on $[0,1]$). I have problems with the proof that every continuous functional on this space is zero. I.e. there are no nontrivial continuous linear functionals on this (metric) space.

Answer 1

We use the following

Proposition: Let $E$ be a topological vector space (not necessarily Hausdorff) over $\mathbb{R}$ or $\mathbb{C}$. There exists a nonzero continuous linear functional on $E$ if and only if there is a convex neighbourhood of $0$ that isn't the whole space.

Proof: Let $\lambda \neq 0$ be a continuous linear functional on $E$. Then $U = \lambda^{-1}(\{ z : \lvert z\rvert < 1\})$ is a convex and balanced proper subset of $E$. Since $\{ z : \lvert z\rvert < 1\}$ is open (in $\mathbb{R}$ or $\mathbb{C}$, as the case may be), and $\lambda$ is continuous, it follows that $U$ is open, and since $0\in U$, we see that $0$ has a nontrivial [meaning not the entire space] convex neighbourhood of $0$.

Conversely, suppose $V\neq E$ is a convex neighbourhood of $0$ in $E$. The Minowski functional

$$\mu_V \colon x \mapsto \inf \{ t > 0 : x \in tV\}$$

is then a sublinear functional (since $V$ is convex). Choose $x_0 \in E\setminus V$. By the Hahn-Banach extension lemma, we can find a real-linear functional $\lambda \colon E \to \mathbb{R}$ with $\lambda(x_0) = \mu_V(x_0)$ and $\lambda(x) \leqslant \mu_V(x)$ for all $x\in E$. Then $V \subset \lambda^{-1}\bigl((-1,1)\bigr)$, which shows that $\lambda$ is continuous. If $E$ is an $\mathbb{R}$-vector space, we're done, and if it's a complex vector space, let $\tilde{\lambda}(x) = \lambda(x) - i\lambda(ix)$ be the complex linear functional with real part $\lambda$. Clearly $\tilde{\lambda}$ is also continuous, and $\tilde{\lambda} \neq 0$.

---

Hence, to show that on a topological vector space no nonzero continuous linear functional exists, it suffices to show that the only convex neighbourhood of $0$ is the whole space. Equivalently, the convex hull of every neighbourhood of $0$ is the whole space.

Now, in the space $\mathscr{M}([0,1])$ of measurable functions on $[0,1]$, endowed with the pseudometric

$$d(f,g) = \int_0^1 \frac{\lvert f(x) - g(x)\rvert}{1 + \lvert f(x) - g(x)\rvert}\,dx$$

(we get a proper metric if we divide out the space of functions vanishing almost everywhere, but that's not necessary to do), let $V$ a neighbourhood of $0$. By definition of the topology, there is an $\varepsilon > 0$ such that

$$B_{\varepsilon}(0) = \{ f \in \mathscr{M}([0,1]) : d(f,0) < \varepsilon\} \subset V.$$

Choose $n \in \mathbb{N}$ such that $\frac{1}{n} < \varepsilon$. For $g \in \mathscr{M}([0,1])$ and $0 \leqslant k < n$, let

$$g_k(x) = \begin{cases} n\cdot g(x) &, \frac{k}{n} \leqslant x < \frac{k+1}{n} \\ \quad 0 &, \text{ otherwise} \end{cases}$$

(for $k = n-1$, let $g_{n-1}(x) = n\cdot g(x)$ on the closed interval $\bigl[\frac{n-1}{n},1\bigr]$ rather than on the half-open interval). Then we have

$$d(g_k,0) = \int_0^1 \frac{\lvert g_k(x)\rvert}{1 + \lvert g_k(x)\rvert}\,dx = \int_{k/n}^{(k+1)/n} \frac{n\lvert g(x)\rvert}{1 + n\lvert g(x)\rvert}\,dx < \int_{k/n}^{(k+1)/n} 1\,dx = \frac{1}{n} < \varepsilon,$$

and hence

$$g = \sum_{k = 0}^{n-1} \frac{1}{n}\cdot g_k$$

belongs to the convex hull of $B_{\varepsilon}(0)$.

Thus $\operatorname{conv} V = \mathscr{M}([0,1])$ for every neighbourhood of $0$, which by the proposition above implies that there is no nonzero continuous linear functional on $\mathscr{M}([0,1])$.

Answer 2

This is Problem 51(vi) of Chapter 14 in Real Analysis, fourth edition, by Royden and Fitzpatrick. Minkowski functionals are not introduced at that point in the text. Rather, the authors give the following hint:

Let $\psi: X\to\mathbf R$ be linear and continuous where $X$ is your space. Show that there is a real number $b$ such that $\psi(c) = 0$ whenever $c$ is the characteristic function of an interval of length less than $1/b$. Hence, $\psi(h) = 0$ for all step functions $h$.

Let $c$ be the characteristic function of an interval $(a,a + 1/b]$ contained in $[0,1]$. If $a = 0$, read $(a,a + 1/b]$ as $[0,1/b]$. Because $\psi$ is continuous, there is a $\delta$ for which if $\rho(\lambda c,0) <\delta$, then $|\psi(\lambda c) - 0| < 1$ where $\lambda$ is an nonzero real number. Now $$\rho(\lambda c,0) =\int_0^1\frac{|\lambda c - 0|}{1 +|\lambda c - 0|} =\int_a^{a+1/b}\frac{|\lambda c|}{1 +|\lambda c|} <\int_a^{a+1/b} 1 =\frac1b.$$ Choose $b$ large enough so that $\rho(\lambda c,0) <\delta$. In that case, $|\psi(\lambda c)| < 1$, or $|\psi(c)| < 1/|\lambda|$ by linearity of $\psi$. Because that holds for $|\lambda|$ arbitrarily large, $\psi(c) = 0$. Thus, there is a $b$ such that $\psi(c) = 0$ whenever $c$ is the characteristic function of an interval of length less than $1/b$.

Step functions are linear combinations of such characteristic functions; hence, $\psi(h) = 0$ for all step functions $h$ by linearity of $\psi$.

We next show that the set of step functions on $[0,1]$ is dense in $X$. Take a nonempty open subset of $X$. Because the subset is nonempty, it contains a function $f$ of $X$. And because the subset is open, it contains an open ball centered at $f$ of radius $\epsilon$. If necessary, reduce $\epsilon$ so that $0 <\epsilon < 3$. By Lusin's theorem, there is a continuous function $g$ on $\mathbf R$ and a closed set $F$ contained in $[0,1]$ for which $$f = g\text{ on $F$ and }m([0,1]\sim F) <\epsilon/3.$$ Set $F$ is nonempty because $\epsilon < 3$ and bounded because $F\subseteq [0,1]$. Then $g$ is bounded on $F$ by the extreme value theorem. Moreover, $g$ is measurable on $F$ because it equals measurable function $f$ on $F$ (or because $g$ is continuous on $F$). By Problem 18 of Chapter 3, there is a step function $h$ on $F$ and a measurable subset $A$ of $F$ for which $$|h - f| <\epsilon/3\text{ on $A$ and }m(F\sim A) <\epsilon/3.$$ Then\begin{align} \rho(f,h) & =\int_0^1\frac{|f - h|}{1 +|f - h|}\\ & <\int_{[0,1]\sim F} 1 +\int_{F\sim A} 1 +\int_A\frac\epsilon3\\ & <\epsilon/3 +\epsilon/3 +\epsilon/3\\ & =\epsilon,\end{align} which shows that $h$ is contained in the open ball and therefore in the open subset. Hence, the set of step functions on $[0,1]$ is dense in $X$.

We have shown that $\psi$ takes the same values as the zero functional on a dense subset of metric space $X$. Hence, $\psi = 0$ on $X$. Thus, there are no nonzero continuous linear functionals on $X$.