Let be the Lebesgue measure on . Consider , the space of Lebesgue measurable functions for which there exists some constant such that for every , . 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 (one can actually show that the infimum can be attained). For , define . One can show that is a topological vector space with the local base .

If were locally convex, then there would exist and a convex open neighborhood of such that .

We consider the function , where . Note that

for all . Therefore .

Now, for each , define

,

where is a “mod 1 translate” of . That is, for , and for .

Note that

- each satisfies (same argument as before), and so as it is a convex combination of elements in ;
- on each interval , is strictly decreasing (you can verify this by differentiation);
- for all (that is to say is “periodic”).

Therefore,

.

Observe that for each , one has

.

The first inequality holds because as we remarked before the infimum can be attained, and the second inequality comes from . In particular, this holds for . So something of order is bounded above by , which is impossible.

Therefore, cannot be locally convex.