Scalar series as a constant Hilbert profile

By [A1] and [F7], the assumed AC yields both DC and Countable Choice, so the product-extension setup in [F1] is available. Use the constant-profile map $\mathcal{I}:\mathbb{R}\to H$ defined in the Example. Then $\mathrm{\parallel }\mathcal{I}(y){\mathrm{\parallel }}_H^2={\int }_0^1\mathrm{\mid }y{\mathrm{\mid }}^{2}d\xi =\mathrm{\mid }y{\mathrm{\mid }}^2$, so $\mathcal{I}$ is a linear isometry. Since $\mathcal{I}$ is continuous, $X_t(\omega )=\mathcal{I}(Y_t(\omega ))$ is a measurable causal $H$-valued shift; applying $\mathcal{I}$ to the same independent replacements gives $X_{t,m}(\omega )=\mathcal{I}(Y_{t,m}(\omega ))$.

The isometry gives $\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H^2=\mathbb{E}\mathrm{\mid }{Y}_0{\mathrm{\mid }}^2$ and $(\mathbb{E}\mathrm{\parallel }{X}_0-X_{0,m}{\mathrm{\parallel }}_H^2{)}^{1\mathrm{/}2}=(\mathbb{E}\mathrm{\mid }{Y}_0-Y_{0,m}{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}$ for every $m$. Apply [F2] to $Y_0$ and the constant random variable $1$ to obtain $\mathbb{E}\mathrm{\mid }{Y}_0\mathrm{\mid }\le (\mathbb{E}\mathrm{\mid }{Y}_0{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$; hence $\mathbb{E}\mathrm{\parallel }{X}_0{\mathrm{\parallel }}_H<\mathrm{\infty }$. By [A1] and [F7], AC supplies Countable Choice, so [F1] gives a countable dense sequence $(h_k)$ in $H$. For each $N$, let ${\pi }_N(x)$ be a nearest point among $h_1,\dots ,h_N$, resolving ties by the least index. Its finitely many cells are Borel, since they are finite intersections of strict or weak comparisons of continuous distance functions. Thus ${\pi }_N\circ X_0$ is a measurable finite-valued simple map, and density gives ${\pi }_N(X_0)\to X_0$ pointwise. By [F6], $X_0$ is strongly measurable, so [F3] gives its Bochner integrability. Since [F3] also gives scalar simple functions $s_j={\sum }_k{c}_{j,k}{\mathbf{1}}_{A_{j,k}}$ with $\mathbb{E}\mathrm{\mid }{s}_j-Y_0\mathrm{\mid }\to 0$, their images satisfy ${\mathbb{E}}_B\mathcal{I}(s_j)={\sum }_k\mathbb{P}(A_{j,k})\mathcal{I}(c_{j,k})=\mathcal{I}(\mathbb{E}{s}_j)$. The isometry implies $\mathbb{E}\mathrm{\parallel }\mathcal{I}(s_j)-X_0{\mathrm{\parallel }}_H\to 0$, so taking the defining integral limits gives ${\mathbb{E}}_B{X}_0=\mathcal{I}(\mathbb{E}{Y}_0)=0$. The profile shift therefore has the required centering and exactly the scalar replacement-error sum.

The functional $L$ is well-defined on almost-everywhere classes, and [F5] gives $\mathrm{\mid }L(v)\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_H\mathrm{\parallel }1{\mathrm{\parallel }}_H=\mathrm{\parallel }v{\mathrm{\parallel }}_H$. Since $L(1)=1$ and $\mathrm{\parallel }1{\mathrm{\parallel }}_H=1$, $\mathrm{\parallel }L\mathrm{\parallel }=1$; also $L(X_t)={\int }_0^1{Y}_{t}d\xi =Y_t$. All hypotheses of [F4] are therefore satisfied, so its conditional bootstrap conclusion applies to this scalar summary.