= Solution
Identify $G$ with the <Boolean hypercube> $\{-1,1\}^d$, with coordinatewise multiplication and normalized <Haar measure> assigning mass $2^{-d}$ to each point. Its <Bernoulli function (hypercube)> $\epsilon_i(\omega)=\omega_i$ is a <Rademacher random variable>. The <Walsh functions on a hypercube> are the <Walsh characters> $w_A=\prod_{i\in A}\epsilon_i$, indexed by all subsets $A\subseteq\{1,\ldots,d\}$, including $w_\varnothing=1$. Independence of the sign coordinates gives $\langle w_A,w_B\rangle=\mathbb E w_{A\triangle B}=\mathbf1_{A=B}$. There are $2^d$ characters, equal to the dimension of $L^2(G)$, so they form an <orthonormal basis>.
Flipping coordinate $i$ negates $w_A$ exactly when $i\in A$. Thus the <coordinate-flip generator on a hypercube> satisfies
$$
\boxed{Lw_A=-|A|w_A,\qquad\sigma(L)=\{0,-1,\ldots,-d\},\qquad\text{multiplicity of }-j=\binom dj.}
$$
For $f=\sum_A\widehat f(A)w_A$, <Parseval identity> gives $\langle f,Lf\rangle=-\sum_A|A||\widehat f(A)|^2\leq0$. Equivalently its <Dirichlet form of a Markov chain> is
$$
\mathcal E(f)=-\mathbb E[fLf]=\frac14\sum_{i=1}^d\mathbb E\bigl(f(\omega^{(i)})-f(\omega)\bigr)^2,
$$
where $\omega^{(i)}$ flips coordinate $i$.
Set $S(\omega)=\sum_i a_i\epsilon_i(\omega)$ and $F(\omega)=\|S(\omega)\|$. Since $F(-\omega)=F(\omega)$, its Walsh expansion contains only sets $A$ of even size. Every nonconstant such set has $|A|\geq2$, giving the <even-function spectral gap on a hypercube>
$$
2\operatorname{Var}(F)\leq\mathcal E(F).
$$
At each $S(\omega)$ the <Hahn-Banach theorem> supplies a real supporting linear functional $\ell$ of <norm> at most one with $\ell(S)=\|S\|$; for a complex normed space use the real part of a complex norming functional. At $S=0$ take $\ell=0$. Convexity of the <norm> gives $F(\omega^{(i)})-F(\omega)\geq\ell(S(\omega^{(i)})-S(\omega))$. Since $LS=-S$, sum these inequalities to get $LF\geq-F$. Therefore $\mathcal E(F)=-\mathbb E[FLF]\leq\mathbb EF^2$. Combining the two bounds yields the <sharp Rademacher second-moment inequality>
$$
\boxed{\mathbb E\left\|\sum_i a_i\epsilon_i\right\|^2\leq2\left(\mathbb E\left\|\sum_i a_i\epsilon_i\right\|\right)^2.}
$$
No smoothness of the <norm> is required. The constant $2$ is sharp: two equal nonzero real coefficients give modulus $0$ or $2|a|$ with equal probabilities. This is the second-versus-first moment case of the <Kahane-Khintchine inequality> for arbitrary <normed vector spaces>.
For the complex-circle assertion, write each independent <Steinhaus random variable> as $\eta_i=\epsilon_i\cos\theta_i+i\delta_i\sin\theta_i$, where $\theta_i$ is uniform on $[0,\pi/2]$ and all quadrant signs $\epsilon_i,\delta_i$ are independent. Conditional on the angles, $\sum_i a_i\eta_i$ is a <Rademacher sum> with $2d$ complex coefficients $a_i\cos\theta_i$ and $ia_i\sin\theta_i$. Its conditional second moment is $\sum_i|a_i|^2$, independent of the angles. The preceding inequality gives its conditional first moment at least $\sqrt{\frac12\sum_i|a_i|^2}$. Average over the angles and square to obtain the <Steinhaus first-moment lower bound>
$$
\boxed{\sum_i|a_i|^2\leq2\left(\mathbb E\left|\sum_i a_i\eta_i\right|\right)^2.}
$$
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