Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 201 / 4

Write $A_n={\max }_{0\le k<2^n}|{\xi }_{(k+1)2^{-n}}-{\xi }_{k{2}^{-n}}|$. The maximum of nonnegative numbers to the power $p$ is bounded by their sum, so the increment hypothesis givesTherefore $\Vert A_n{\Vert }_p\le C{2}^{-n(\beta -1/p)}$. The Minkowski inequality and completeness of $L^p$ imply convergence of the defining series wheneverFor a fully explicit argument, the norm of every tail is at most $2C{\sum }_{n\ge N}{2}^{-n(\beta -1/p-\alpha )}$, which tends to zero. Since the summands are nonnegative, their pointwise increasing sum equals this finite $L^p$ limit and is finite almost surely. This is a guaranteed range; no endpoint convergence follows from the displayed summability estimate. The assumptions control increments only and do not require ${\xi }_0\in L^p$.

Fix an exponent in the range from (a), and work on the single probability-one event where $K_{\alpha }<\infty$ and all dyadic values are finite. We give the dyadic increment chaining argument. For dyadic $s<t$, put $\delta =t-s$ and choose $n\ge 0$ with $2^{-n}\le \delta <2^{1-n}$. Let $s_j=2^{-j}\lfloor 2^{j}s\rfloor$ and $t_j=2^{-j}\lfloor 2^{j}t\rfloor$. The level-$n$ approximations are at most two grid steps apart, so their difference is bounded by $2A_n$. At each subsequent level an approximation either stays fixed or moves by one adjacent level increment. Because $s,t$ are dyadic, these approximations eventually equal $s,t$. ThusThe dyadic rationals are dense in $[0,1]$, so this uniform bound gives a unique continuous extension $X$ to the entire interval, withAll these equalities hold on that one event, not merely one event per dyadic time. Define $X$ to be the zero path on its null complement. Each $X_t$ is measurable as the limit of the random variables at deterministic left dyadic approximations. Alternatively, their measurable linear interpolations converge uniformly to $X$, which also shows measurability as a random element of $C([0,1])$. This is the continuous extension version of a continuous modification.

The Gaussian process assumption and the covariance formula give a centered Gaussian increment with varianceFor every finite $p>1$, a standard normal variable $Z$ therefore givesFor any $0<\alpha <1/2$, choose $p>\max (2,(1/2-\alpha {)}^{-1})$. Then $\alpha <1/2-1/p$, so parts (a) and (b) give a Hölder continuous function of exponent $\alpha$ as an extension. To obtain one extension with every required exponent, take a countable sequence ${\alpha }_j\uparrow 1/2$ and apply the construction with suitable $p_j$. Intersect the probability-one events. Their continuous extensions agree on the dense set $D$ and hence agree everywhere. For each smaller exponent choose $j$ with $\alpha <{\alpha }_j$; since $|t-s|\le 1$, the bound with ${\alpha }_j$ also implies the bound with $\alpha$. ThusThe exponent constants may depend on $\alpha$ and the sample path. This proves Brownian Hölder regularity for the continuous Gaussian extension without making an uncountable intersection of unrelated full-probability events.