*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.