Sorry in advance if this is not sufficiently research-level, it is really more of a reference request since the proof is not difficult. Let $\mathcal{Y}$ be a compact set, let $\{X_n\}$ denote a sequence of random variables, and let $f(x,y)$ and $g(y)$ be "nice" functions. Suppose that for each fixed $y\in\mathcal{Y}$, we have $$\liminf_{n\to\infty} f(X_n,y)\geq g(y)$$ almost surely. What is the appropriate theorem to cite that says that (for sufficiently nice functions), we have $$\liminf_{n\to\infty} \min_{y\in\mathcal{Y}} f(X_n,y)\geq \min_{y\in\mathcal{Y}} g(y)$$ almost surely?


1 Answer 1


First here, without loss of generality (wlog) $g=0$ (otherwise, replace $f(x,y)$ and $g(y)$ by $f(x,y)-g(y)$ and $0$, respectively). Second, in view of the almost-sure (a.s.) condition and the a.s. desired conclusion, wlog each random variable $X_n$ takes only one value, say $x_n$ -- which let us assume for simplicity to be a real number.

So, the desired result takes this simplified form:

$$\liminf_{n\to\infty} f(x_n,y)\ge0 \tag{1}$$ for some "nice" function $f$ and all $y$ in a compact set $Y:=\mathcal Y$, then $$\liminf_{n\to\infty}\min_{y\in Y} f(x_n,y)\ge0. \tag{2}$$

Consider now the following example: Let $Y:=[-1,1]$ and let $f(x,y):=-xy^2\,e^{-xy^2}$ for real $x$ and $y\in Y$. Probably everyone will agree that this function $f$ is nice. Let $x_n\to\infty$ (as $n\to\infty$). Then $f(x_n,y)\to0$ for each $y\in Y$, so that (1) holds. However, $\min_{y\in Y} f(x_n,y)=-1/e$ for each large enough $n$ (namely, for each $n$ such that $x_n\ge1$), so that (2) fails to hold.

The obvious reason for the just described phenomenon is that the sequence $(x_n)$ is unbounded.

So, it is natural to assume that $|x_n|\le M$ for some real $M>0$ and all $n$. It is also natural to assume that the "nice" function $f$ is continuous. So, $f$ is uniformly continuous on the compact set $[-M,M]\times Y$. Also, passing to a subsequence, wlog assume that $x_n\to x_\infty$ for some real $x_\infty$.

So, $f(x_n,y)\to f(x_\infty,y)$ uniformly in $y\in Y$ and hence $$\lim_{n\to\infty}\min_{y\in Y} f(x_n,y) =\min_{y\in Y} f(x_\infty,y) =\min_{y\in Y}\lim_{n\to\infty}f(x_n,y)\ge0, $$ by (1). So, we have deduced, very simply, (2) from (1).

Essentially, all we used here is that any function continuous on a compact set $K$ is uniformly continuous on $K$. So, one can hardly expect a reference involving something more substantial than this.


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