Proof of Uniform Mean-Square Continuity on a Compact Interval

If $a=b$, uniformity is trivial (any $\delta$ works, the only pair being $s=t=a$), and $t\mapsto\mathbb{E}[H_t^{2}]$ takes a single finite value, hence is bounded. So assume $a<b$.

Step 1 (Uniformity). Suppose, for contradiction, that there is $\varepsilon_0>0$ such that for every natural number $k\ge1$ there are $s_k,t_k\in[a,b]$ with $|s_k-t_k|<1/k$ and $\lVert H_{s_k}-H_{t_k}\rVert_{2}\ge\varepsilon_0$. The sequence $(s_k)$ is a bounded sequence, so by the Bolzano-Weierstrass theorem there is a subsequence $(s_{k_i})_i$ with limit $u$; since $a\le s_{k_i}\le b$ for all $i$, we get $u\in[a,b]$ (if $u<a$, taking $\varepsilon=a-u$ in the definition of the limit would force $s_{k_i}<a$ for large $i$; similarly $u\le b$). From $|s_{k_i}-t_{k_i}|<1/k_i\to0$ we get $t_{k_i}\to u$ as well.

By mean-square continuity at $u$, choose $\delta>0$ with $\lVert H_v-H_u\rVert_{2}<\varepsilon_0/2$ for all $v\in[a,b]$ with $|v-u|<\delta$. For $i$ large enough, $|s_{k_i}-u|<\delta$ and $|t_{k_i}-u|<\delta$, so by the triangle inequality for the mean-square norm (Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm)

contradicting $\lVert H_{s_{k_i}}-H_{t_{k_i}}\rVert_{2}\ge\varepsilon_0$. This proves uniform mean-square continuity.

Step 2 (Continuity and boundedness of the second moment). For $s,t\in[a,b]$, the triangle inequality of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm applied to $H_t=(H_t-H_s)+H_s$ and to $H_s=(H_s-H_t)+H_t$ gives

Hence $t\mapsto\lVert H_t\rVert_{2}$ is continuous on $[a,b]$, and so is its square $t\mapsto\mathbb{E}[H_t^{2}]=\lVert H_t\rVert_{2}^{2}$ by claim 4 (continuity arithmetic) of Sum and Product Rules for One-Dimensional Derivatives and Continuity. A continuous function on the closed bounded interval $[a,b]$ with $a<b$ is bounded by the extreme value theorem. $\square$