= Solution
Use the same finite truncation $T_m(h)=\sum_{j\leq m}\langle h,e_j\rangle A_j$ for $f$ and $g$. Since $\mathbb E[A_iA_j]=\delta_{ij}$ by <independence>, zero <means> and unit <variances>,
$$
\mathbb E[T_m(f)T_m(g)]
=\sum_{j\leq m}\langle f,e_j\rangle\langle g,e_j\rangle.
$$
Here $\delta_{ij}$ is the <Kronecker delta>. The <Cauchy-Schwarz inequality> and <convergence in L2> justify passing to the limit in the product's <expectation>. The <Parseval identity for a Hilbertian basis> now gives
$$
\boxed{\mathbb E[(Wf)(Wg)]=\langle f,g\rangle_{\mathcal H}
=\int_0^1f(t)g(t)\,dt.}
$$
Because each $Wh$ has <mean> zero, this is also its <covariance> identity. Thus $W$ is an <isonormal Gaussian process> over $\mathcal H$.
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