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We have functions $f_n\in L^1$ such that $\int f_ng$ has a limit for every $g\in L^\infty$. Does there exist a function $f\in L^1$ such that the limit equals $\int fg$? I think this is not true in general (really? - why?), then can this be true if we also know that $f_n$ belong to a certain subspace of $L^1$?

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up vote 8 down vote accepted

Another way of phrasing your question: Is $L^{1}$ weakly sequentially complete? That is to say: does every weak Cauchy sequence in $L^1$ converge?

The answer is yes.

(Added: See Nate's answer for a definition of weak Cauchy sequence and why this "completeness" of $L^1$ may be considered surprising).

Here's an outline of the argument (I'm assuming that you're working on a $\sigma$-finite measure space $(\Omega, \Sigma, \mu)$, but that's not an essential restriction, see the end of this answer):

  • First of all, it follows from the uniform boundedness theorem that the sequence $(f_{n})$ is bounded in $L^{1}$.

  • Next, we can define for a measurable set $E$ a quantity $$\nu(E) = \lim_{n \to \infty} \int_{E} f_{n}$$ because the characteristic function of $E$ belongs to $L^{\infty}$.

  • Then one can verify that $\nu$ is a (signed) measure which is absolutely continuous with respect to $\mu$. By the Radon-Nikodym theorem it follows that $\nu(E) = \int_{E} f\,d\mu$ for a unique function (class) $f \in L^{1}$.

  • Now by definition we have $\displaystyle \int f g = \lim_{n \to \infty} \int f_{n} g$ for all characteristic functions $g$. But as the characteristic functions span a dense subspace of $L^{\infty}$ we conclude that this must hold for all $g \in L^{\infty}$.

You can find the details of this argument e.g. in Dunford-Schwartz, Linear operators I, Theorem IV.8.6 on page 290. The assumption on $\sigma$-finiteness I made is easily removed, as is also explained there: The point is that the union of the supports of the $f_n$ is $\sigma$-finite, so we may assume that we work on a $\sigma$-finite space in the first place.

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Thank you so much. – peno May 23 '11 at 21:00
How do we verify that $\nu$ is countably additive? Why can we switch lim and sum in $\sum_{m=1}^{\infty} \lim_{n \to \infty} \int_{E_m} f_{n}=\lim_{n \to \infty} \sum_{m=1}^{\infty} \int_{E_m} f_{n}$ when $E_m$'s are mutually disjoint? – Ti Wen Nov 29 '15 at 18:58
Secondly, let $(g_m)$ be a norm convergent simple functions in $L^\infty$ with limit $g$, then we need to verify: $$\int f g=\lim_{m \to \infty}\int f g_m =\lim_{m \to \infty}\lim_{n \to \infty} \int f_{n} g_m=\lim_{n \to \infty}\lim_{m \to \infty} \int f_{n} g_m=\lim_{n \to \infty} \int f_{n} g~~?$$ The first and the last equalities follow from DCT. The second one is also easy to verify, but I don't know why the third equality holds. These details are not included in Dunford-Schwartz's book~~~ – Ti Wen Nov 29 '15 at 19:20

Perhaps surprisingly, the answer is yes.

More generally, given any Banach space $X$, a sequence $\{x_n\} \subset X$ is said to be weakly Cauchy if, for every $\ell \in X^*$, the sequence $\{\ell(f_n)\} \subset \mathbb{R}$ (or $\mathbb{C}$) is Cauchy. If every weakly Cauchy sequence is weakly convergent, $X$ is said to be weakly sequentially complete.

Every reflexive Banach space is weakly sequentially complete (a nice exercise with the uniform boundedness principle). $L^1$ is not reflexive, but it turns out to be weakly sequentially complete anyway. This theorem can be found in P. Wojtaszczyk, Banach spaces for analysts, as Corollary 14 on page 140. It works for $L^1$ over an arbitrary measure space.

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Many thanks to you too. – peno May 23 '11 at 21:03

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