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Suppose I have a sequence of random continuous functions, $f^{n} : [0, t] \to \mathbb{R}$. Suppose there also exists a random continuous function, $f: [0, t] \to \mathbb{R}$, defined on the same probability space.

I know that $f^{n}$ converges (almost surely) pointwise to $f$. I also know that $f^{n}$ converges in distribution to $f$ as an element of the space of continuous functions from $[0,t]$ to $\mathbb{R}$, where this convergence is with respect to the topology of uniform convergence. Can I then conclude that $f^{n}$ converges almost surely uniformly to $f$?

I have been trying to use tightness conditions described in Kallenbergs Foundations of Modern Probability, but haven't been able to prove the statement, which makes me think it might be false? Thanks

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  • $\begingroup$ What does it mean to say $f^n$ converges in distribution to $f$? $\endgroup$ Commented Dec 1 at 6:36
  • $\begingroup$ @AnthonyQuas If F is bounded functional from $C[0,t] \to \mathbb{R}$ continuous wrt. to the topology of uniform convergence on $C[0,t]$ then it is the case that: $$\lim\limits_{n \to \infty} \mathbb{E}(F(f^{n})) = \mathbb{E}(F(f))$$ $\endgroup$
    – Snidd
    Commented Dec 1 at 7:08
  • $\begingroup$ I’m just checking: by functional, you mean linear functional, right? $\endgroup$ Commented Dec 2 at 23:05

1 Answer 1

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No. Take $f^n$ all independent to be $0$ with probability $1/\sqrt{n}$ and consisting of a bump of height $1$ and width $1/n$ at a uniformly distributed location otherwise. Clearly $f^n \to 0$ in law (and in probability) and, for each $x$, $f^n(x) \to 0$ almost surely by Borel-Cantelli, but $\sup f^n$ doesn't converge to $0$ by the second Borel-Cantelli lemma.

You can however conclude that $f^n \to f$ in probability (in the uniform topology). This is an instance for example of Proposition 3.12 in this article of mine (but I am sure this was known before).

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