[proofplan]
The proof is the Hilbert-space orthogonality computation in expectation. We expand the squared norm of the random signed sum using the [inner product](/page/Inner%20Product) on $H$, separate diagonal terms from cross terms, and then use the Rademacher moment identities together with independence. The diagonal terms contribute exactly $\|x_i\|_H^2$, while every cross term has expectation zero. The stated Rademacher type $2$ inequality follows by taking square roots of the resulting identity.
[/proofplan]
[step:Define the random signed sum and expand its squared Hilbert norm]
Let $(\cdot,\cdot)_H$ denote the inner product on $H$, linear in the first argument if $H$ is complex. Define the $H$-valued [random variable](/page/Random%20Variable)
\begin{align*}
S:\Omega &\to H
\end{align*}
by
\begin{align*}
S(\omega)=\sum_{i=1}^n \varepsilon_i(\omega)x_i.
\end{align*}
For each $\omega\in\Omega$, the Hilbert norm is induced by the inner product, so
\begin{align*}
\|S(\omega)\|_H^2=(S(\omega),S(\omega))_H.
\end{align*}
Expanding the finite inner product sum gives
\begin{align*}
\|S(\omega)\|_H^2=\sum_{i=1}^n\sum_{j=1}^n \varepsilon_i(\omega)\varepsilon_j(\omega)(x_i,x_j)_H.
\end{align*}
In the complex case this double sum is real because it is equal to $(S(\omega),S(\omega))_H$; equivalently, it may be written as the diagonal contribution plus real parts of paired off-diagonal terms.
[/step]
[step:Compute the expectation of each coefficient $\varepsilon_i\varepsilon_j$]
For each $i\in\{1,\ldots,n\}$, the Rademacher moment identity [citetheorem:10110] gives
\begin{align*}
\mathbb E[\varepsilon_i^2]=1.
\end{align*}
If $i,j\in\{1,\ldots,n\}$ and $i\neq j$, then $\varepsilon_i$ and $\varepsilon_j$ are independent. Since each is bounded and real-valued, their product is integrable, and independence gives
\begin{align*}
\mathbb E[\varepsilon_i\varepsilon_j]=\mathbb E[\varepsilon_i]\mathbb E[\varepsilon_j].
\end{align*}
Again by [citetheorem:10110],
\begin{align*}
\mathbb E[\varepsilon_i]=0
\end{align*}
and
\begin{align*}
\mathbb E[\varepsilon_j]=0.
\end{align*}
Therefore
\begin{align*}
\mathbb E[\varepsilon_i\varepsilon_j]=0
\end{align*}
whenever $i\neq j$.
[guided]
We need to understand which terms in the inner product expansion survive after taking expectation. The coefficients in that expansion are the scalar random variables $\varepsilon_i\varepsilon_j$.
First consider the diagonal case $i=j$. For each $i\in\{1,\ldots,n\}$, $\varepsilon_i$ is a [Rademacher random variable](/page/Rademacher%20Random%20Variable), so the moment computation in [citetheorem:10110] gives
\begin{align*}
\mathbb E[\varepsilon_i^2]=1.
\end{align*}
This is the source of the diagonal contribution $\|x_i\|_H^2$.
Now consider the off-diagonal case $i\neq j$. The hypothesis says that the family $\varepsilon_1,\ldots,\varepsilon_n$ is independent, so the two random variables $\varepsilon_i$ and $\varepsilon_j$ are independent. Since Rademacher random variables only take the values $-1$ and $1$, the product $\varepsilon_i\varepsilon_j$ is bounded and hence integrable. The factorisation property of expectation for independent integrable random variables gives
\begin{align*}
\mathbb E[\varepsilon_i\varepsilon_j]=\mathbb E[\varepsilon_i]\mathbb E[\varepsilon_j].
\end{align*}
The same Rademacher moment identity [citetheorem:10110] gives
\begin{align*}
\mathbb E[\varepsilon_i]=0
\end{align*}
and
\begin{align*}
\mathbb E[\varepsilon_j]=0.
\end{align*}
Thus
\begin{align*}
\mathbb E[\varepsilon_i\varepsilon_j]=0.
\end{align*}
This is the cancellation mechanism: independence turns the expectation of the product into a product of expectations, and the zero mean of each Rademacher variable kills every cross term.
[/guided]
[/step]
[step:Take expectation and cancel the cross terms]
The function $\omega\mapsto \|S(\omega)\|_H^2$ is bounded by the deterministic finite quantity
\begin{align*}
\left(\sum_{i=1}^n \|x_i\|_H\right)^2,
\end{align*}
because $|\varepsilon_i(\omega)|=1$ for every $i$ and every $\omega$ outside a null set. Hence it is integrable. By linearity of expectation over the finite double sum,
\begin{align*}
\mathbb E[\|S\|_H^2]=\sum_{i=1}^n\sum_{j=1}^n \mathbb E[\varepsilon_i\varepsilon_j](x_i,x_j)_H.
\end{align*}
Using the coefficient computation from the previous step, all terms with $i\neq j$ vanish, while the terms with $i=j$ satisfy $\mathbb E[\varepsilon_i^2]=1$. Therefore
\begin{align*}
\mathbb E[\|S\|_H^2]=\sum_{i=1}^n (x_i,x_i)_H.
\end{align*}
Since $(x_i,x_i)_H=\|x_i\|_H^2$ for each $i$, we obtain
\begin{align*}
\mathbb E\left[\left\|\sum_{i=1}^n \varepsilon_i x_i\right\|_H^2\right]=\sum_{i=1}^n \|x_i\|_H^2.
\end{align*}
[/step]
[step:Deduce Rademacher type $2$ with constant $1$]
The definition of Rademacher type $2$ with constant $T$ requires
\begin{align*}
\left(\mathbb E\left[\left\|\sum_{i=1}^n \varepsilon_i x_i\right\|_H^2\right]\right)^{1/2}\leq T\left(\sum_{i=1}^n \|x_i\|_H^2\right)^{1/2}
\end{align*}
for every $n\in\mathbb N$, every $x_1,\ldots,x_n\in H$, and every independent Rademacher family $\varepsilon_1,\ldots,\varepsilon_n$. The identity already proved gives equality with $T=1$:
\begin{align*}
\left(\mathbb E\left[\left\|\sum_{i=1}^n \varepsilon_i x_i\right\|_H^2\right]\right)^{1/2}=\left(\sum_{i=1}^n \|x_i\|_H^2\right)^{1/2}.
\end{align*}
Thus $H$ has Rademacher type $2$ with type constant $1$.
[/step]
