 Solution

First use independent fair signs ${\epsilon }_i\in \{-1,1\}$, the Rademacher distribution convention for Bernoulli random variables in this setting. Let $\Omega =\{-1,1{\}}^d$ with uniform probability, write $s(\omega )={\sum }_i{\omega }_i{a}_i$ and $F(\omega )=\Vert s(\omega )\Vert$. We prove the sharp Rademacher second-moment inequality by a complete finite-cube calculation.

For a function on this cube define the coordinate-flip generator on a hypercubewhere ${\omega }^{(i)}$ flips just coordinate $i$. The Walsh characters ${\chi }_A(\omega )={\prod }_{i\in A}{\omega }_i$ form an orthonormal basis: distinct characters have expectation-zero product, and there are $2^d$ of them. Flipping a coordinate in $A$ changes its sign and other flips leave it unchanged, so $L{\chi }_A=-|A|{\chi }_A$. Expand the real function $F$ as ${\sum }_A\hat{F}(A){\chi }_A$. Since $F(-\omega )=F(\omega )$, changing variables $\omega \mapsto -\omega$ shows that $\hat{F}(A)=0$ whenever $|A|$ is odd. ThusThis derives the even-function spectral gap on a hypercube, rather than assuming it.

At each vertex with $s(\omega )\ne 0$, choose a norming functional ${\ell }_{\omega }$ of norm one satisfying ${\ell }_{\omega }(s(\omega ))=F(\omega )$, supplied by the Hahn-Banach theorem. For complex $E$ take real parts in the following inequalities. At a vertex with $s(\omega )=0$, use the zero functional. The supporting inequality for the norm givesBut $Ls=-s$, since every coefficient is a degree-one Walsh character. Summing the preceding inequalities with factor $1/2$ therefore proves $LF(\omega )\ge -F(\omega )$. As $F\ge 0$,Rearranging and taking square roots yieldsNo completeness of $E$ is needed. The constant is sharp: with two identical nonzero scalar coefficients, the absolute sum is zero with probability $1/2$ and twice the coefficient magnitude with probability $1/2$.

If Bernoulli instead means fair $\{0,1\}$ variables $b_i$, write $b_i=(1+{\epsilon }_i)/2$ and $a={\sum }_i{a}_i$. Add an independent fair sign ${\epsilon }_0$ and apply the proved estimate to $Y={\epsilon }_{0}a+{\sum }_i{\epsilon }_i{a}_i$. Conditional on either sign of ${\epsilon }_0$, $\Vert Y\Vert$ has the same distribution as $\Vert a+{\sum }_i{\epsilon }_i{a}_i\Vert$, by symmetry of the sign sum. Hence $\Vert Y{\Vert }_{L^p(E)}=2\Vert {\sum }_i{b}_i{a}_i{\Vert }_{L^p(E)}$ for $p=1,2$, proving the same bound in that convention.

The constant is documented in the primary sharp-constant study.

Fairness is essential. For a general $\{0,1\}$ Bernoulli random variable with success probability $0<t<1/2$, a single nonzero scalar coefficient gives $L^2/L^1=1/\sqrt{t}>\sqrt{2}$. Thus the printed term cannot be read as permitting arbitrary Bernoulli probabilities.