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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.

Petra Axolotl
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2 Answers2

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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).

Mittens
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  • I completely get it that if a quasi-norm is not locally convex, the dual space must be trivial. What I think is still missing is that $L^{\infty,1}(R)$ is not locally convex. (I have not studied topology. In fact, local convexity is so far not even mentioned in the textbook I am using. So an elementary proof would be much appreciated!) – Petra Axolotl Nov 16 '22 at 22:01
  • The example $f_N$ shows that $L^{\inty,1}$ is not, for $f_N$ is a convex combination of functions in the ball $B(0;2)$. – Mittens Nov 16 '22 at 22:14
  • I understand. My question is whether the existence of such $f_N$ necessarily means that every other function $f$ is also a convex combination of functions in a neighbourhood the origin. In the other proof of yours about $L^p, p<1$, it was explicitly shown to be valid for all $f$. – Petra Axolotl Nov 17 '22 at 16:41
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    @PetraAxolotl: To show that the space is not locally convex is enough to consider one neighborhood of the form $B(0;r)$ and show that there are elements in that neighborhood say, $f_1,\ldots,f_m$ such that a (not every, but just some) convex combination $f:=\alpha_1 f_1+\ldots + \alpha_m f_m$ ($0<\alpha_j<1$, $\sum^m_{j=1}\alpha_j=1$) is not in $B(0;r)$. – Mittens Nov 17 '22 at 17:13
  • I see. Thanks a lot! – Petra Axolotl Nov 17 '22 at 19:37
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There are examples of such spaces which do have non trivial linear functionals. For example the discrete version of $L^p$, i.e. $\ell^p$. Point evaluations do the trick. (If you want a silly example, take a finite dimensional space and put a suitable quai-norm on it).

lake
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