I'm reading the book Theoretical Statistics by Keener, and I couldn't figure out one of the claims in the proof for Lemma 9.1.
Lemma 9.1 states: let $W$ be a random function in $C(K)$ where $K \subset \mathbb{R}^p$ is compact. Define $\mu(t)=EW(t)$, $t \in K$ to be the mean of $W$. If $E ||W||_{\infty} < \infty$, then $\mu$ is continuous. Also, define $M_\epsilon(t) = \sup_{s:||s-t||<\epsilon} |W(s)-W(t)|$, then
$$\sup_{t \in K} E M_\epsilon(t) \to 0 \quad \text{as } \epsilon \to 0.$$
In the proof for the second claim, it's stated that:
- $M_\epsilon$ is continuous because $W$ is continuous.
- By continuity, $M_\epsilon(t) \to 0$ as $\epsilon \to 0$.
Here's my understanding. The first statement says that for each $\epsilon>0$, $M_\epsilon$ is continuous in $t$. This is needed in the later part of the proof to show $\lambda_\epsilon = E M_\epsilon(t)$ is continuous. I don't know how to prove this statement, would appreciate any comments!!
The second statement says that for each $t$, $M_\epsilon(t)$ is continuous in $\epsilon$ at $\epsilon=0$. I can prove this using the fact that $W$ is uniformly continuous on $K$, thanks to the comment from @Mushu Nrek. Here's a sketch of the proof: I want to show for any positive sequence $\{\epsilon_n\}$ where $\epsilon_n \to 0$, $M_{\epsilon_n}(t) \to M_0(t)=0$. Now fix any $\gamma>0$, by uniform continuity of $W$, there exists $\delta>0$ such that $M_\delta(t)<\gamma$. Find $N$ large such that $\epsilon_N < \delta$, now $|M_{\epsilon_n}(t)|<|M_{\epsilon_N}(t)|<\gamma$ for all $n>N$. Hence completing the proof.
I think that there are two things.
First concerning the proof of the Lemma. I don't think that one needs to show that $M_\epsilon$ is continuous in $t$. Indeed, since $W$ is continuous on a compact, it is uniformly continuous. Therefore, independently of $t \in K$, there exists for every $\delta > 0$ some $\epsilon_0 > 0$ such that $M_\epsilon(t) < \delta$ for all $\epsilon < \epsilon_0$. In particular, $\sup_{t\in K} M_\epsilon(t) < \delta$. Then, the result follows by dominated convergence.
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But now answering your question: Why is $M_\epsilon$ continuous in $t$? Let $t_n \to t$ in $K$. Writing $B(r,\epsilon)$ for the ball around $r$ with radius $\epsilon$, we conclude that for every $s\in B(t,\epsilon)$, there exists some $s_n\in B(t_n, \epsilon)$ such that $\Vert s - s_n\Vert \leq \Vert t - t_n\Vert$. By the uniform continuity of $W$ on $K$, we may thus for every $\delta > 0$ choose $N$ large enough such that for every $n\geq N$, there is for every $s\in B(t, \epsilon)$ some $s_n\in B(t_n, \epsilon)$ with $$ \vert W(s_n) - W(s)\vert < \delta. $$ In particular, $$ \vert W(s) - W(t)\vert \leq \vert W(s_n) - W(t_n)\vert + \vert W(s_n) - W(s)\vert + \vert W(t_n) - W(t)\vert $$ (and similarly in the opposite direction) and therefore $$ M_\epsilon(t_n) - 2\delta \leq M_\epsilon(t) \leq M_\epsilon(t_n) + 2\delta $$ for all $n\geq N$. Since $\delta$ was arbitrary, this means that $M_\epsilon(t_n) \to M_\epsilon(t)$ as $n\to +\infty$.