# $f_n$ converges in duality with $C_b$ and is uniformly bounded then it converges in duality with $L^1$

Let $(X,m)$ be a metric measure space, $(f_n)_n$ a sequence in $L^\infty, f \in L^\infty$ s.t. $$\int gf_n \ dm \rightarrow \int g f$$ for every $g : X \rightarrow \mathbb R$ continuous and bounded. I have seen in a book the claim that if $\forall n, \|f_n\|_{L^\infty} \leq C$ then $$\int hf_n \ dm \rightarrow \int h f ,$$ for every $h \in L^1$.

How can I prove this result?

This is true provided that $C_b \cap L^1$ is dense in $L^1$, which holds if $m$ is a finite measure, or if $X$ is locally compact and $m$ is Radon (I imagine many other conditions are possible). In this case, we can use a standard triangle inequality trick. Given any $h \in L^1$ and any $\epsilon > 0$ we can find $g \in C_b \cap L_1$ such that $\|g-h\|_{L^1} < \epsilon$. Then \begin{align*}\left| \int f_n h\,dm - \int f h\,dm \right| &= \left| \int (f_n h - f_n g + f_n g - f g + f g - f h)\,dm \right| \\ &\le \left| \int f_n (h-g)\,dm \right| + \left| \int (f_n - f) g\,dm \right| + \left| \int f (g-h)\,dm \right| \end{align*} The first term is bounded by $\|f_n\|_{L^\infty} \|g-h\|_{L^1} < C \epsilon$. The second term goes to 0 by assumption. The third term is bounded by $\|f\|_{L^\infty} \|g-h\|_{L^1} < \|f\|_{L^\infty} \epsilon$. Hence $$\limsup_{n \to \infty} \left| \int f_n h\,dm - \int f h\,dm \right| \le C \epsilon + \|f\|_{L^\infty} \epsilon + \limsup_{n \to \infty} \left| \int (f_n - f) g\,dm \right| = C \epsilon + \|f\|_{L^\infty} \epsilon.$$ Since $\epsilon$ was arbitrary the limsup on the left must be 0. It is a nonnegative sequence, so it must actually converge to 0, which is what was desired.
I am not exactly sure what can happen if $C_b \cap L^1$ is not dense in $L^1$.
• Thank you! However this does not work in spaces where continuous functions are not dense in $L^1$, but I guess they are too ugly anyway... – Brainstorming Jun 3 '15 at 12:47
• @Brainstorming: Thanks for the correction. I missed that we were working on an arbitrary metric space - I saw an $m$ and assumed it was Lebesgue measure. – Nate Eldredge Jun 3 '15 at 13:50