## Weak L^1 is not locally convex

Let $\mu$ be the Lebesgue measure on $[0,1]$. Consider $L^{1,\infty}[0,1]$, the space of Lebesgue measurable functions $f$ for which there exists some constant $C=C(f)>0$ such that for every $\alpha>0$, $\mu(\{x: |f(x)|>\alpha\})\leq C/\alpha$. The purpose of this post is to show that this is not a locally convex topological vector space (it should be a well known result but it is hard to find it on google, so I write a note here). For topological vector space, it just means a vector space with a topology so that the addition and scalar multiplication are continuous. You may read for example Rudin’s functional analysis for more details.

Let $\Delta(f) = \inf\{C>0: \mu(\{x: |f(x)|>\alpha\})\leq C/\alpha \; \forall\alpha>0\}$ (one can actually show that the infimum can be attained). For $r>0$, define $B_r = \{f\in L^{1,\infty}[0,1] :\Delta(f) < r \}$.  One can show that $L^{1,\infty}[0,1]$ is a topological vector space with the local base $\{B_r:r>0\}$.

If $L^{1,\infty}[0,1]$ were locally convex, then there would exist $0 and a convex open neighborhood $V$ of $0$ such that $B_{r_0} \subset V \subset B_{r_1}$.

We consider the function $f(x)=1/x$, where $x\in (0,1)$. Note that

$\mu(\{x:1/x>\alpha\}) = \min\{1,\alpha^{-1}\}$

for all $\alpha>0$. Therefore $\Delta(f)=1$.

Now, for each $n\in\mathbb{N}$, define

$\displaystyle g_n(x) = \frac{1}{n} \sum_{i=1}^n \frac{r_0}{2}f_{i/n}(x)$,

where $f_{i/n}$ is a “mod 1 translate” of $f$. That is, $f_{i/n}(x) = 1/(x-i/n)$ for $i/n, and $f_{i/n}(x) = 1/(x+1-i/n)$ for $0.

Note that

• each $f_{i/n}$ satisfies $\Delta(f_{i/n}) = 1$ (same argument as before), and so $g_n\in V$ as it is a convex combination of elements in $B_{r_0}$;
• on each interval $(i/n, (i+1)/n)$, $g_n$ is strictly decreasing (you can verify this by differentiation);
• $g_n(x + 1/n (\text{mod }1)) = g_n(x)$ for all $x \in (0,1)$ (that is to say $g_n$ is “periodic”).

Therefore,

$\displaystyle\begin{array}{rl} \inf_{x\in(0,1)} g_n(x) = & \inf_{x\in(0,1/n)} g_n(x) \\ =& \lim_{x\rightarrow 1/n^{-}} \frac{r_0}{2n}\left(\frac{1}{x} + \frac{1}{x+1-\frac{1}{n}} + \frac{1}{x+1-\frac{2}{n}} + \cdots + \frac{1}{x+1-\frac{n-1}{n}} \right) \\ = & \frac{r_0}{2n} \left( n + 1 + \frac{n}{n-1} + \frac{n}{n-2} + \cdots +n \right) \\ \approx & \log{n}\end{array}$.

Observe that for each $\alpha>0$, one has

$\alpha\mu(\{x: |g_n(x)|>\alpha\}) \leq \Delta(g_n) .

The first inequality holds because as we remarked before the infimum can be attained, and the second inequality comes from $V\subset B_{r_1}$. In particular, this holds for $\alpha = \inf_{x\in(0,1)} g_n(x)$. So something of order $\log{n}$ is bounded above by $r_1$, which is impossible.

Therefore, $L^{1,\infty}[0,1]$ cannot be locally convex.