Second moment bound for a finite martingale-difference sum

For each coordinate $0\le \mathrm{\ell }<d$, $Z_j^{\mathrm{\ell }}\in L^2$ implies $Z_j^{\mathrm{\ell }}\in L^1$ on the probability space, since $2\mathrm{\mid }z\mathrm{\mid }\le z^2+1$ and the space has total mass one. Thus the scalar supplier [F2] defines $M_j^{\mathrm{\ell }}=\mathbb{E}[Z_j^{\mathrm{\ell }}\mid {\mathcal{F}}_{j-1}]$. This gives the coordinatewise vector $M_j$. The assumption ${\xi }_j\in L^2$ and the identity $M_j=Z_j-{\xi }_j$ imply $M_j\in L^2$. By scalar linearity and the fact that an integrable measurable variable is its own conditional expectation, each coordinate satisfies $\mathbb{E}[{\xi }_j^{\mathrm{\ell }}\mid {\mathcal{F}}_{j-1}]=0$. Integrating the scalar conditional second-moment bound and using [F2] gives $\mathbb{E}\mathrm{\parallel }{\xi }_j{\mathrm{\parallel }}_2^2\le \mathbb{E}[v_j]\mathrm{.}$ Thus the vector conditional means, errors and second moments used below are well-defined in the stated finite-dimensional $L^2$ spaces.

If $i<j$, then [given] ${\xi }_i$ is ${\mathcal{F}}_{j-1}$-measurable. For each coordinate $\mathrm{\ell }$, both ${\xi }_i^{\mathrm{\ell }}$ and ${\xi }_j^{\mathrm{\ell }}$ are in $L^2$; the pointwise inequality $2\mathrm{\mid }ab\mathrm{\mid }\le a^2+b^2$ makes their product integrable. Apply [F3] coordinatewise and use step 1.1 to obtain $\mathbb{E}⟨{\xi }_i,{\xi }_j⟩={\sum }_{\mathrm{\ell }<d}\mathbb{E}[{\xi }_i^{\mathrm{\ell }}{\xi }_j^{\mathrm{\ell }}]={\sum }_{\mathrm{\ell }<d}\mathbb{E}[{\xi }_i^{\mathrm{\ell }}\mathbb{E}[{\xi }_j^{\mathrm{\ell }}\mid {\mathcal{F}}_{j-1}]]=0.$ Expanding the finite squared norm and using the integrated conditional variance bound from step 1.1 gives $\mathbb{E}{\parallel {\sum }_{j=1}^m{\xi }_j\parallel }_2^2={\sum }_{j=1}^m\mathbb{E}\mathrm{\parallel }{\xi }_j{\mathrm{\parallel }}_2^2\le {\sum }_{j=1}^m\mathbb{E}[v_j]\mathrm{.}$

Since $Z_j=M_j+{\xi }_j$, apply $\mathrm{\parallel }u+v{\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2+2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$ to the sums $u={\sum }_j{M}_j$ and $v={\sum }_j{\xi }_j$. Take expectations and substitute step 2.1. This proves the Statement.