*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. —**

*The set $E$ is a vector subspace of the space of all measurable functions (modulo coincidence almost everywhere).**The mapping $d$ is a distance on $E$.**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.

Thank you for this post, very informative for an undergraduate like myself.

ReplyDeleteOut of curiosity, is research in topological vector spaces alive and well or are most of the (interesting) questions already answered?