关于Keener《Theoretical Statistics》引理9.1证明中$M_\epsilon(t)$关于$t$连续性的疑问

Hey there, let's unpack why $M_\epsilon(t)$ is continuous in $t$ for fixed $\epsilon > 0$—this is a tricky little detail, so it's totally normal to get stuck here!

First, let's recap the key definitions to align our context:

• $K \subset \mathbb{R}^p$ is compact, and $W$ is a random function in $C(K)$, meaning for every sample point, $W(\cdot)$ is a continuous function on $K$ (and since $K$ is compact, $W(\cdot)$ is actually uniformly continuous on $K$ too—we’ll lean on that later).
• $M_\epsilon(t) = \sup_{s: |s - t| < \epsilon} |W(s) - W(t)|$

To prove $M_\epsilon(t)$ is continuous at some $t_0 \in K$, we need to show that for any $\gamma > 0$, there exists a $\delta > 0$ such that if $|t - t_0| < \delta$, then $|M_\epsilon(t) - M_\epsilon(t_0)| < \gamma$. We can split this into two complementary parts: bounding the limsup of $M_\epsilon(t)$ as $t \to t_0$, and bounding the liminf.

Step 1: $\limsup_{t \to t_0} M_\epsilon(t) \leq M_\epsilon(t_0)$ #

Take any $t$ close to $t_0$, say $|t - t_0| < \delta$ (we’ll pick $\delta$ strategically later). For any $s$ with $|s - t| < \epsilon$, the triangle inequality gives:
$$|W(s) - W(t)| \leq |W(s) - W(t_0)| + |W(t_0) - W(t)|$$
If we take the supremum over all such $s$, we get:
$$M_\epsilon(t) \leq \sup_{|s - t| < \epsilon} |W(s) - W(t_0)| + |W(t) - W(t_0)|$$

Now choose $\delta < \epsilon/2$: this means $|s - t| < \epsilon$ implies $|s - t_0| \leq |s - t| + |t - t_0| < \epsilon + \delta < 3\epsilon/2$. Since $W$ is uniformly continuous on $K$, we can make $\delta$ small enough that:

• $\sup_{|s - t_0| < 3\epsilon/2} |W(s) - W(t_0)| < M_\epsilon(t_0) + \gamma/2$
• $|W(t) - W(t_0)| < \gamma/2$

Combining these gives $M_\epsilon(t) < M_\epsilon(t_0) + \gamma$ for all $t$ with $|t - t_0| < \delta$, so the limsup can’t exceed $M_\epsilon(t_0)$.

Step 2: $\liminf_{t \to t_0} M_\epsilon(t) \geq M_\epsilon(t_0)$ #

Now let's reverse the logic. Take any $s$ with $|s - t_0| < \epsilon$. If we pick $\delta < \epsilon - |s - t_0|$ (positive because $|s - t_0| < \epsilon$), then $|t - t_0| < \delta$ implies $|s - t| \leq |s - t_0| + |t_0 - t| < \epsilon$. Using the triangle inequality again:
$$|W(s) - W(t_0)| \leq |W(s) - W(t)| + |W(t) - W(t_0)|$$
Taking the supremum over all such $s$ gives:
$$M_\epsilon(t_0) \leq M_\epsilon(t) + |W(t) - W(t_0)|$$

As $t \to t_0$, $|W(t) - W(t_0)| \to 0$, so rearranging gives $M_\epsilon(t) > M_\epsilon(t_0) - \gamma$ for all sufficiently close $t$ to $t_0$. This means the liminf is at least $M_\epsilon(t_0)$.

Putting it all together #

Since the limsup and liminf both equal $M_\epsilon(t_0)$, we have $\lim_{t \to t_0} M_\epsilon(t) = M_\epsilon(t_0)$—so $M_\epsilon(t)$ is continuous at $t_0$. Since $t_0$ was arbitrary, $M_\epsilon(t)$ is continuous on all of $K$.

Also, huge props to you for working out the continuity of $M_\epsilon(t)$ in $\epsilon$ using uniform continuity—your proof sketch is perfect, that's exactly how you leverage compactness of $K$ to make that argument work!

备注：内容来源于stack exchange，提问作者statstats