On Lp-spaces, when 0<p<1, convex sets and linear forms

While the theory of normed vector spaces is now extensively taught at the undergraduate level, the more general theory of topological vector spaces usually does not reach the curriculum. There may be good reasons for that, and here is an example, taken from a paper of Mahlon M. Day, The spaces $L^p$ with $0<p<1$ (Bull. Amer. Math. Soc. 46 (1940), 816–823), of which I learned from a nice analysis blurb by Keith Conrad which has almost the same title.

For simplicity, I consider here the simple case when the measured space is $[0;1]$, with the Lebesgue measure, and $p=1/2$. Let $E$ be the set of measurable real valued functions $f$ on the interval $[0;1]$ such that $\int_0^1|f(t)|^{1/2}dt<+\infty$, where we identify two functions which coincide almost everywhere. For $f,g\in E$, let us define $d(f,g)=\int_0^1 \mathopen|f(t)-g(t) \mathclose|^{1/2}dt$.

Lemma. —

1. The set $E$ is a vector subspace of the space of all measurable functions (modulo coincidence almost everywhere).
2. The mapping $d$ is a distance on $E$.
3. With respect to the topology defined by $d$, the addition of $E$ and the scalar multiplication are continuous, so that $E$ is a topological vector space.

Proof. — We will use the following basic inequality: For $u,v\in\mathbf R$, one has $\mathopen|u+v\mathclose|^{1/2}\leq |u|^{1/2}+|v|^{1/2}$; it can be shown by squaring both sides of the inequality and using the usual triangular inequality. Let $f,g\in E$; taking $u=f(t)$ and $v=g(t)$, and integrating the inequality, we obtain that $f+g\in E$. It is clear that $af\in E$ for $a\in\mathbf R$ and $f\in E$. This proves that $E$ is a vector subspace of the space of measurable functions. For $f,g\in E$, one has $f-g\in E$, so that $d(f,g)$ is finite. Let then $f,g,h\in E$; taking $u=f(t)-g(t)$ and $v=g(t)-h(t)$, and integrating this inequality for $t\in[0;1]$, we then obtain the triangular inequality $d(f,h)\leq d(f,g)+d(g,h)$ for $d$. Moreover, if $d(f,g)=0$, then $f=g$ almost everywhere, hence $f=g$ by definition of $E$. This proves that $d$ is a distance on $E$. Let us now show that $E$ is a topological vector space with respect to the topology defined by $d$. Let $f,g\in E$. For $f',g'\in E$, one then has $d(f'+g',f+g)=\int_0^1\mathopen|(f-f')+(g-g')\mathclose|^{1/2}\leq d(f,f')+d(g,g')$. This proves that addition is continuous on $E$. Similarly, let $a\in \mathbf R$ and $f\in E$. For $b\in\mathbf R$ and $g\in E$, one has $d(af,bg)\leq d(af,bf)+d(bf,bg)\leq \mathopen|b-a\mathclose|^{1/2} d(f,0)+|b|^{1/2}d(f,g)$. This implies that scalar multiplication is continuous. QED.


The following theorem shows one unusual feature of this topological vector space.

Theorem. — One has $E^*=0$: every continuous linear form on $E$ vanishes identically.

Proof. — Let $\phi$ be a non-zero continuous linear form on $E$. Let $f\in E$ be such that $\phi(f)\neq 0$; we may assume that $\phi(f)\geq 1$. For $s\in[0,1]$, let $g_s\colon[0;1]\to\mathbf R$ be the function defined by $g_s(t)=0$ for $0\leq t\leq s$ and $g_s(t)=1$ for $s< t\leq 1$. When $s$ goes from $0$ to $1$, $d(g_s f,0)$ goes from $d(f,0)$ to $0$. Consequently, there exists $s$ such that $d(g_s f,0)=d(f,0)/2$. Then $d((1-g_s)f,0)=\int_0^s |f(t)|^{1/2}dt=\int_0^1|f(t)|^{1/2}dt-\int_s^1|f(t)|^{1/2}dt=d(f,0)-d(g_sf,0)=d(f,0)/2$ as well. Moreover the equality $1=\phi(f)=\phi(g_sf)+\phi((1-g_s)f)=0$ shows that either $\phi(g_sf)\geq1/2$ or $\phi((1-g_s)f)\geq 1/2$. Set $f'=2g_s f$ in the first case, and $f'=2(1-g_s)f$ in the latter; one has $\phi(f')\geq 1$ and $d(f',0)=d(f,0)/\sqrt 2$. Iterating, we obtain a sequence $(f^{(n)})$ of elements of $E$ which converges to $0$ but such that $\phi(f^{(n)})\geq 1$ for every $n$, contradicting the continuity of $\phi$. QED.


On the other hand, we may believe to remember the Hahn-Banach theorem according to which, for every non-zero function $f\in E$, there exists a continuous linear form $\phi\in E^*$ such that $\phi(f)=1$. Obviously, the previous theorem seems to violate the Hahn-Banach theorem.
So why is this not so? Precisely because the Hahn-Banach theorem makes the fundamental hypothesis that the topological vector space be a normed vector space or, more generally, a locally convex vector space, which means that $0$ admits a basis of convex neighborhoods. According to the following proposition, this is far from being so.

Proposition. — $E$ is the only non-empty convex open subset of $E$.

Proof. — Let $V$ be a non-empty convex open subset of $E$. Up to an affine transformation, in order prove that $V=E$, we may assume that $0\in V$ and that $V$ contains the unit ball of center $0$. We first show that $V$ is unbounded. For every $n\geq 1$, we split the interval $[0,1]$ in $n$ intervals $[(k-1)/n,k/n]$, for $1\leq k\leq n$, with characteristic functions $g_k$. One has $d(n^2g_k,0)=1$ for every $k$, hence $n^2 g_k\in V$; moreover, $1=\sum_{k=1}^n g_k$, so that $n=\frac 1n \sum_{k=1}^n n^2 g_k$ belongs to $V$. More generally, given $f\in E$ and $n\geq 1$, we split the interval $[0;1]$ into $n$ successive intervals, with characteristic functions $g_k$, such that $d(fg_k,0)=d(f,0)/n$ for every $k$; one also has $f=\sum fg_k$. Then $d(nfg_k,0)=\sqrt n d(fg_k,0)=1/\sqrt n\leq 1$, hence $n fg_k\in V$ and the relation $f=\frac1n \sum nf g_k$ shows that $f\in V$. QED.



When $(X,\mu)$ is a measured space and $p$ is a real number such that $0<p<1$, the space $L^p(X,\mu)$ has similar properties. For this, I refer the interested reader to the above cited paper of Day and to Conrad's note.

Radon measures form a sheaf for a natural Grothendieck topology on topological spaces

First post of the year, so let me wish all of you a happy new year!

Almost two years ago, Antoine Ducros and I released a preprint about differential forms and currents on Berkovich spaces. We then embarked in revising it thoroughly; unfortunately, we had to correct a lot of inaccuracies, some of them a bit daunting. We made a lot of progress and we now have a much clearer picture in mind. Fortunately, all of the main ideas remain the same.

A funny thing emerged, which I want to explain in this blog.

One of our mottos was to define sheaves of differential forms, or of currents. Those differential forms were defined in two steps : by definition, they are locally given by tropical geometry, so we defined a presheaf of tropical forms, and passed at once to the associated sheaf. What we observed recently is that it is worth spending some time to study the presheaf of tropical forms.

Also, Grothendieck topologies play such an important rôle in analytic geometry over non-archimedean fields; this is obvious for classical rigid spaces, but they are also important in Berkovich geometry, in particular if you want to care about possibly non-good spaces for which points may not have a neighborhood isomorphic to an affinoid space. So it was natural to sheafify the presheaf of tropical forms for the G-topology, giving rise to a G-sheaf of G-forms.

Now, every differential form of maximal degree $\omega$ on a Berkovich space $X$ gives rise to a measure on the topological space underlying $X$. Our proof of this is a bit complicated, and was made more complicated by the fact that we first tried to define the integral $\int_X \omega$, and then defined $\int_X f\omega$ for every smooth function $f$, and then got $\int_X f\omega$ for every continuous function with compact support $f$ by approximation, using a version of the Stone-Weierstrass theorem in our context.

In the new approach, we directly concentrate on the measure that we want to construct. For G-forms, this requires to glue measures defined locally for the G-topology. As it comes out (we finished to write down the required lemmas today), this is quite nice.

Since Berkovich spaces are locally compact, we may restrict ourselves to classical measure theory on locally compact spaces. However, we may not make any metrizability assumption, nor any countability assumption, since the most basic Berkovich spaces lack those properties. Assume that the ground non-archimedean field $k$ is the field $\mathbf C((t))$ of Laurent series over the field $\mathbf C$ of complex numbers. Then the projective line $\mathrm P^1$ over $k$ is not metrizable, and the complement of its ``Gauss point'' $\gamma$ has uncountably many connected components (in bijection with the projective line over $\mathbf C$). Similarly, the complement of the Gauss point in the projective plane $\mathrm P^2$ over $k$ is connected, but is not countable at infinity, hence not paracompact.

As always, there are two points of view on measure theory: Borel measures (countably additive set functions on the $\sigma$-algebra of Borel sets) and Radon measures (linear forms on the vector space of continuous compactly supported functions). By the theorem of Riesz, they are basically equivalent: locally finite, compact inner regular Borel measures are in canonical bijection with Radon measures. Unfortunately, basic litterature is not very nice on that topic; for example, Rudin's book constructs an outer regular Borel measure which may not be inner regular, while for us, the behavior on compact sets is really the relevant one.

Secondly, we need to glue Radon measures defined on the members of a G-cover of our Berkovich space $X$. This is possible because Radon measures on a locally compact topological space naturally form a sheaf for a natural Grothendieck topology!

Let $X$ be a locally compact topological space and let us consider the category of locally compact subspaces, with injections as morphisms.  Radon measures can be restricted to a locally compact subspace, hence form a presheaf on that category.

Let us decree that a family $(A_i)_{i\in I}$ of locally compact subspaces of a locally compact subspace $U$ is a B-cover (B is for Borel) if for every point $x\in U$, there exists a finite subset $J$ of $I$ such that $x\in A_i$ for every $i\in J$ and such that $\bigcup_{i\in J}A_i$ is a neighborhood of $x$. B-covers form a G-topology on the category of locally compact subsets, for which Radon measures form a sheaf! In other words, given Radon measures $\mu_i$ on members $A_i$ of a B-cover of $X$ such that the restrictions to $A_i\cap A_j$ of $\mu_i$ and $\mu_j$ coincide, for all $i,j$, then there exists a unique Radon measure on $X$ whose restriction to $A_i$ equals $\mu_i$, for every $i$.

This said, the proof (once written down carefully) is not a big surprise, nor specially difficult,  but I found it nice to get a natural instance of sheaf for a Grothendieck topology within classical analysis.