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<r_0<r_1 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<x\leq 1, and f_{i/n}(x) = 1/(x+1-i/n) for 0<x<i/n.

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”).


\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) <r_1.

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.

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