 Solution

We use independent symmetric Rademacher random variables ${\epsilon }_j\in \{-1,1\}$, the intended meaning of the Bernoulli variables in this context. Without this convention the assertion need not hold: for one $0/1$ Bernoulli variable of success probability $t<1/2$ and one nonzero vector, the proposed inequality would be $\sqrt{t}\le \sqrt{2}t$, which is false.

Let $\Omega =\{-1,1{\}}^n$ with the uniform probability measure, put $S(\epsilon )={\sum }_j{\epsilon }_j{x}_j$, and set $F(\epsilon )=\Vert S(\epsilon )\Vert$. We prove the sharp estimate rather than assuming the Kahane-Khintchine inequality. If ${\epsilon }^{(j)}$ denotes $\epsilon$ with its $j$th sign reversed, defineThe Walsh characters $w_A(\epsilon )={\prod }_{j\in A}{\epsilon }_j$ form an orthonormal basis of scalar functions on this finite cube, and $\mathcal{L}{w}_A=-|A|w_A$. Since $F(-\epsilon )=F(\epsilon )$, its coefficients for odd $|A|$ vanish. Every nonconstant remaining character has $|A|\ge 2$. Expanding in this orthonormal basis gives the even-function spectral gap on a hypercubeAt each $\epsilon$, the Hahn-Banach theorem supplies a real norming linear functional $\varphi$ with $\Vert \varphi \Vert \le 1$ and $\varphi (S(\epsilon ))=F(\epsilon )$. For a complex Banach space use its underlying real space. Since $\Vert z\Vert \ge \varphi (z)$,Adding gives $\mathcal{L}F\ge -F$. Multiplying by the nonnegative $F$ yields $F(-\mathcal{L}F)\le F^2$. ThereforeThe constant is sharp: take $n=2$ and $x_1=x_2\ne 0$ in the scalar field. The random norm is $0$ or $2\Vert x_1\Vert$, each with probability one half.

A Banach space has cotype 2 if a constant $C$, independent of the length of the family, satisfiesFor $f_j\in L^1(0,1)$, the triangle inequality for the Euclidean norm givesFor real or complex scalars $a_j$, independence gives $\mathbb{E}|{\sum }_j{\epsilon }_j{a}_j{|}^2={\sum }_j|a_j{|}^2$. The proved sharp Rademacher second-moment inequality, applied pointwise to $(f_j(t))$, and Fubini theorem now yieldThus $L^1(0,1)$ has cotype 2, with $C\le \sqrt{2}$.