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Let $F_n, F : [0, 1] \to \mathbb{R}$ be absolutely continuous (with respect to the Lebesgue measure). Assume that $F_n \to F$ weakly in the sense that $\int_0^1 g(t) dF_n(t) \to \int_0^1 g(t) dF(t)$ for all continuous bounded functions $g : [0, 1] \to \mathbb{R}$. (By the absolute continuity, the measures $dF_n$ and $dF$ have a Lebesgue density $f_n$ and $f$, i.e. $dF_n = f_n dt$ and $dF = f dt$ and we have weak convergence of the measures $dF_n = f_n dt \to f dt = dF$.)

Absolutely continuous functions are continuous and so they can be considered as elements in $C[0,1]$. Does it follow that $F_n \to F$ uniformly, that is $\sup_{t \in [0,1]} |F_n(t) - F(t)| \to 0$ ?

What if we add some uniform boundedness condition like $\sup_n \sup_t |f_n(t)| < \infty$?

shashlik
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1 Answers1

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As stated, the answer is "no." Consider a continuous version of the typewriter sequence, which converges to $0$ weakly. The functions $F_n$ of this sequence are all uniformly bounded by $1$ even, but it's plain to see $F_n(x)$ does not converge to $0$ for any $x\in [0,1]$, let alone uniform convergence.

It is true however that $F_n\to 0$ in measure in this example, from which it is possible to extract a subsequence converging pointwise, and hence almost uniformly. Beyond this, however, additional assumptions should be thought of for the sequence $F_n$ to satisfy.

Alex Ortiz
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