I was having some trouble with the following problem from Luenberger's Optimization by Vector Space Methods (p. 72).
Update (2/7/2021): I attempted (a) based on a hint from GReyes and gave another shot at (b), updating the solutions below. Please comment if you see any mistakes!
Consider the set $X$ of all real functions $x$ defined on $(-\infty, \infty)$ for which $$\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |x(t)|^2 dt < \infty$$
Let $M$ be the subspace on which the indicated limit is zero.
a) Show that the space $H = X/M$ becomes a pre-Hilbert space when the inner product is defined as $$([x]|[y]|) = \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T x(t)y(t) dt$$
b) Show that $H$ is not separable.
a) I first show that $\langle[x],[y]\rangle$ has the same value regardless of which function is chosen from each coset. Let $x$ and $\tilde{x}$ be arbitrary functions in $[x]$ and $y$ and $\tilde{y}$ arbitrary functions in $[y]$ and consider the difference between each pair's inner products $\langle x, y \rangle – \langle \tilde{x}, \tilde{y}\rangle$. Using the facts that $xy – \tilde{x}\tilde{y} \leq |x||y – \tilde{y}| + |x-\tilde{x}||\tilde{y}|$ and that the integral and limit operations are linear, we obtain the inequality
\begin{gather}
\langle x, y \rangle – \langle \tilde{x}, \tilde{y}\rangle \leq \langle |x|, |y-\tilde{y}|\rangle + \langle |x-\tilde{x}|, |\tilde{y}|\rangle
\text{ and by Cauchy Schwarz, }\\
\langle |x|, |y-\tilde{y}|\rangle + \langle |x-\tilde{x}|, |\tilde{y}|\rangle \leq \lvert|x\rvert| \lvert|y-\tilde{y}\rvert| + \lvert|x – \tilde{x}\rvert| \lvert|y\rvert|
\end{gather}
Because $y – \tilde{y}$ and $x – \tilde{x}$ are in the subspace $M$, $\lvert|y-\tilde{y}\rvert| = 0$ and $\lvert |x – \tilde{x}\rvert | = 0$; and because $\lvert |x \rvert|$ and $\lvert |y \rvert|$ are finite, $\lvert|x\rvert| \lvert|y-\tilde{y}\rvert| + \lvert|x – \tilde{x}\rvert| \lvert|y\rvert| = 0$ and so $\langle x, y \rangle – \langle \tilde{x}, \tilde{y} \rangle = 0$. The elements in the coset were chosen arbitrarily, and so the inequality holds in both directions, meaning that $\langle x, y \rangle – \langle \tilde{x}, \tilde{y} \rangle = 0$ and so $\langle[x],[y]\rangle = \langle[\tilde{x}],[\tilde{y}]\rangle$.
To check whether this is a pre-Hilbert space, I check whether the inner-product satisfies necessary properties when applied to arbitrary functions $x\in [x]$ and $y\in [y]$ from the associated cosets. By the commutativity of multiplication it is clear that $\langle[x],[y]\rangle = \langle[y],[x]\rangle$. From the distributive property of multiplication and linearity of the integral operator, $\langle[x]+[y],[z]\rangle = \langle[x],[z]\rangle + \langle[y],[z]\rangle$ and $\langle\lambda[x],[y]\rangle = \lambda\langle[x],[y]\rangle$. Finally, because $|x(t)|^2 \geq 0$ and $T\to \infty$, the inner product $\langle[x],[x]\rangle \geq 0$. If $\langle[x],[x]\rangle = 0$ then $x\in M$ by construction. If $x\in M$ then its coset includes all $\tilde{x}$ satisfying $\tilde{x} – x \in M$, and so $\tilde{x} \in M$ by the closure of $M$. Satisfying the four necessary conditions, the inner product defines a pre-Hilbert space on $X/M$.
Update (2/7/2021): Here's my attempt at showing that the space is not separable. I assume that $H$ is uncountable. Is this ok? I started writing a proof by trying to show that the inner product is a surjective mapping from any two cosets $[x], [y] \in H$ to $\mathbb{R}$, but I keep having trouble coming up with examples.
b) Proof: If $H$ is separable, it contains a countable dense subset $Z$ and because $H$ is uncountable, $Z$ must be a proper subset. Choose an arbitrary $[g] \in H\setminus Z$ and use $N([g], \delta)$ to denote the neighborhood of radius $\delta$ around $[g]$. Because $Z$ is dense in $H$, for each $\delta > 0$ the intersection $Z \cap N([g], \delta) \neq \emptyset$. But, fixing $[g]$, we can choose $[z] \in N([g], \epsilon/3)$, $\tilde{z}\in [z]$, and $\tilde{g}\in [g]$ so that
\begin{align}
\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} – \tilde{g}|^2 dt &\leq \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} – z|^2dt + \lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |g – \tilde{g}|^2 dt + \epsilon/3
\end{align}
Because $z-\tilde{z}$ and $g-\tilde{g} \in M$, there exists some $T'$ such that for all $T \geq T'$
\begin{align}
\frac{1}{2T} \max\Big(\int_{-T}^T |\tilde{z} – z|^2dt, \int_{-T}^T |g – \tilde{g}|^2 dt\Big) < \epsilon/3
\end{align}
and so
\begin{equation}
\lim_{T\to\infty} \frac{1}{2T} \int_{-T}^T |\tilde{z} – \tilde{g}|^2 dt < \epsilon.
\end{equation}
But then, $\tilde{z} – \tilde{g} \in M$ and so $\tilde{z}$ and $\tilde{g}$ are in the same coset, contradicting the assumption that $[g] \in H\setminus Z$.
Best Answer
To finish your computation you can use Cauchy-Schwartz inequality for both integrals: $$ (\frac 1{2T}\int_{-T}^T|x(t)||\tilde y(t)-y(t)|dt)^2\le \frac 1{2T}\int_{-T}^T|x(t)|^2 dt\cdot\frac 1{2T}\int_{-T}^T|\tilde y(t)-y(t)|^2 dt $$ And a similar computation for the second one. The right hand side goes to zero because one of the limits is zero and the other one is finite by the definitions of the quotient and the space itself.
Checking the Hilbertian structure is straightforward after this.