Existence of non-trivial bounded linear functionals with a quasi-norm

functional-analysisnormed-spaces

Is it well known, and easy to prove, that if we define the norm of $L^p$ with $p<1$ in the usual way, $ \| \cdot \|_{L^p} $ is not actually a norm because it does not satisfy subadditivity. At the same time, if there is a linear functional $l$ that satisfies $$|l(f)| \le M \|f\|_{L^p} $$ for all $f \in L^p$ and some $M > 0$, then $l=0$.

I wonder whether it is generally true that with a quasi-norm $\mathcal{N}$, which violates subadditivity, the only bounded linear functionals is trivial (if we pretend $\mathcal{N}$ is a norm).

In particular, I have the following example in mind (from Exercise 15 in Chapter 2 of Stein and Shakarchi's Functional Analysis).

Consider the weak-type space, consisting of all functions $f$ for which $m(\{ x : |f(x)| > \alpha \}) \leq \frac{A}{\alpha}$ for some $A$ and all $\alpha > 0$. One might hope to define a norm on this space by taking the “norm” of $f$ to be the least $A$ for which the above inequality holds. Denote this quantity by $\mathcal{N}(f)$.

We can show that $\mathcal{N}(f)$ is not a genuine norm because it does not satisfy subadditivity through the following example.

The function $f(x) = 1/ | x |$ has $\mathcal{N}(f) = 2$. But if $f_N = \frac{1}{N} [f(x + 1) + f(x + 2) + \cdots + f(x + N)]$, then $\mathcal{N}(f_N) \ge c \log N$.

The textbook also claims that this space has no non-trivial bounded linear functionals.

Best Answer

The answer to the OP is no, it is not true in general. The key issue is local convexity.

In the context of the OP's posting, a quasi-metric space $X$ is a non empty set $X$ along with a function $\rho:X\times X\rightarrow[0,\infty)$ such that

  1. $\rho(x,y)=0$ iff $x=y$,
  2. $\rho(x,y)=\rho(y,x)$ for all $x,y\in X$,
  3. there is $K\geq1$ with $\rho(x,y)\leq K(\rho(x,z)+\rho(z,y))$ for all $x,y,z\in X$. When $X$ is a vector space with a quasi-metric $\rho$ that is translation invariant ($\rho(x_z,y+z)=\rho(x,y)$ for all $x,y,x\in X$), and homogenous ($\rho(cx,cy)=|c|\rho(x,y)$ for all $x,y\in X$ and $c\in\mathbb{F}$) quasi metric, then $\rho$ is a quasi-norm and $(X,\rho)$ a quasi-normed space.

The quasi-metric $\rho$ induces a topology $\tau(\rho)$ on $X$ which can be construct by uniformities: The sets $U_r=\{(x,y): \rho(x,y)<r\}$ form a basis for a uniformity; in fact the collection $U_{1/n}$, $n\in\mathbb{N}$, is a countable basis for the uniformity, and so $(X,\tau(\rho))$ is metrizable (see, Kelly, J. General Topology, Springer-Verlag, chapter 6).

A more direct result by Macías-Segovia and later by Aimar, Jaffei and Nitti (see Paluszynski, M and Stempak, P., On quasi-metric and metric spaces, Proceedings of the AMS, Vol 137, 12, 2009, pp. 4307-4312) states that

Proposition: If $(X,\rho)$ is a quasi-metric (resp. quais-normed) space, then there is $0<p\leq1$ ($p=\tfrac{\log(2)}{\log(2K)}$) and a metric (resp. norm) $d_p$ on $X$ such that $\tilde{\rho}:=d^{1/p}$ is a quai-metric (resp. quasi-norm) on $X$ equivalent to $\rho$ (i.e., $c\rho\leq\tilde{\rho}\leq d\rho$ for constants $c,d>0$); furthermore, $$\tilde{\rho}(x,y)\leq \Big(\tilde{\rho}^p(x,z)+\tilde{\rho}^p(z,y)\Big)^{1/p},\qquad x,y,z\in X$$

It is clear that the topology induced by the quasi-metric $\rho$ and the metric $d_p$ are the same.

If $X$ is a quasi-normed vector space, ans $d_p$ is complete, the $(X,d_p)$ is an $F$-space. Examples of this situation are already mentioned by the OP: $L_p(X,\mathscr{F},\mu)$ for $0<p<1$ with $\rho(x)=\|x\|_p$ and $L^{\infty,p}(\mathbb{R},\mathscr{B}(\mathbb{R}),m)$ ($m$ Lebeusge measure on the line with $\rho(x)=\sup_{t>0}\{t m^{1/p}(|x|>t))$ with $0<p\leq 1$. These spaces are not normable (there is no norm equivalent to $d_p$) since $L_p$ ($0<p<1$ ) is not (in general) locally convex, and neither is $L^{\infty,1}(\mathbb{R})$. Hence their dual spaces are trivial.

This is in contrast to $L_p(\mu)$ with $p\geq1$ or $L^{\infty,p}(\mu)$ with $p>1$. The former is already a complete normed space; the later has a norm $\newcommand\vertiii[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \vertiii{\;}_p$ that is equivalent to the quasi-norm $[\;]_p$ (see this posting for example).

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