= Solution
A <separable Hilbert space> has a countable subset dense in its <norm topology>.
\b[The printed assertion about all <locally square-integrable functions> is false.] The proposed average is not even finite for every such function: for $f(x)=x$,
$$
\frac1R\int_{-R}^R f(x)^2\,dx=\frac23R^2\longrightarrow\infty.
$$
It also fails positive definiteness. The nonzero function $f=\mathbf1_{[0,1]}$ has
$$
\lim_{R\to\infty}\frac1R\int_{-R}^R f(x)^2\,dx=0.
$$
Consequently this formula cannot define an <inner product>, much less a <Hilbert space>, on $L^2_{\mathrm{loc}}(\mathbb R)$.
A precise version of the intended nonseparability argument uses the <mean-square completion of trigonometric polynomials>. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos(\lambda x)$, and $\sin(\lambda x)$, with arbitrary $\lambda>0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are <orthogonal>, each sine and cosine has squared <norm> one, and the constant function has squared <norm> two. Thus, after collecting equal frequencies,
$$
\left\|a_0+\sum_j\bigl(a_j\cos(\lambda_j x)+b_j\sin(\lambda_j x)\bigr)\right\|^2
=2a_0^2+\sum_j(a_j^2+b_j^2).
$$
This is positive definite on $V$. Its <Hilbert space completion> $H$ contains the uncountable <orthonormal set> $\{\cos(\lambda x):\lambda>0\}$. The distance between two distinct members is $\sqrt2$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: \b[this corrected completed space is nonseparable.] Completion is an essential additional construction; it does not validate the printed claim about all of $L^2_{\mathrm{loc}}$.
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