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Could you help please:

Is there any analogue of continuous mapping theorem for convergence of sequence of random variables in $L^2$?

I mean: If $g$ is a continuous function $\mathbb{R}\to\mathbb{R}$ (not differentiable in general) and $X_n \stackrel{L^2} \to X$ then $g(X_n) \stackrel{L^2} \to g(X).$

Thanks in advance

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  • $\begingroup$ this is tautologically true $\endgroup$ – Norbert Jun 17 '13 at 20:08
  • $\begingroup$ I suppose $g$ is assumed to be a continuous function $\mathbb{R} \to \mathbb{R}$ (not a continuous function on $L^2$ as @Norbert seems to assume) and that $g(X_n)$ denotes composition $g \circ X_n$. Dan, could you please clarify? $\endgroup$ – Martin Jun 18 '13 at 11:00
  • $\begingroup$ Martin, thanks. Surely I assumed $g$ to be continuous $\mathbb{R}→\mathbb{R}$! $\endgroup$ – Dan Jun 18 '13 at 14:24
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Without further assumptions this is false. If $g$ is unbounded, we might not even have $g(X_n), g(X) \in L^2$.

If $g$ is bounded and continuous, then $g(X_n) \to g(X)$ in measure by the continuous mapping theorem, and also in any $L^p$ by the dominated (bounded) convergence theorem. So the statement is true in this case.

It also holds when $g$ is unbounded but Lipschitz: if $C$ is the Lipschitz constant, then $g(X_n), g(X) \in L^2$ because $|g(X_n)| \le |g(0)| + C |X_n|$ where the right side is an $L^2$ random variable. Moreover, we have $$E|g(X_n) - g(X)|^2 \le C^2 E|X_n - X|^2$$ where the latter goes to zero.

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    $\begingroup$ +1. As a side note, there is a section on the book "A Primer in Nonlinear Analysis" by Prodi & Ambrosetti dedicated to those things. Keyword: Nemitsky operators. $\endgroup$ – Giuseppe Negro Jun 18 '13 at 15:03
  • $\begingroup$ Thanks a lot! one last question concerning this problem. Can we say smth about convergence if we know that $g$ - continuous and $g(X_n),g(X)∈L^2,$ but $g$ is not necessarily Lipchitz? $\endgroup$ – Dan Jun 18 '13 at 17:13
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    $\begingroup$ @Dan: From the already mentioned book by Prodi & Ambrosetti, Remark 2.5 pag. 17, it is possible to show that if $g$ is continuous and $g(X)\in L^2$ for all $X\in L^2$, then the map $X\to g(X)$ is continuous with respect to the $L^2$-norm. Unfortunately the proof is not given. $\endgroup$ – Giuseppe Negro Jun 18 '13 at 17:38
  • $\begingroup$ @Giuseppe, thanks! They cite Vainberg, M.M., Variational Methods for the Study of Nonlinear operators. I don't have one. Failed to prove it myself. Apparently I have to ask this question separately from the latter one. $\endgroup$ – Dan Jun 18 '13 at 18:42
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Suppose $g$ is $\alpha$-Hölder continuous, for $\alpha \geq 0$, and that $X_n \overset{L^m}{\to} X$ for some $m > 0$. If $\alpha > 0$, then $g(X_n) \overset{L^{m/\alpha}}{\to} g(X)$. If $\alpha = 0$ (i.e. bounded), and $g$ is continuous, then $g(X_n) \overset{L^m}{\to} g(X)$. The case $\alpha = 1$ corresponds to Lipschitz continuity. The case $\alpha > 0$ is very easy to prove. I'd be interested in more general results.

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