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I have the Hilbert space of square integrable functions on $[a, b]$, and what I would like to have is to discretize this space, i.e., find a sequence of finite-dimensional Hilbert spaces $H_K$ of dimension $K$ such that

$|\langle u, v \rangle - \langle u^{(K)}, v^{(K)} \rangle| \rightarrow 0$

as $K \rightarrow \infty$, where $v^{(K)}$ is a 'discretization' of $v$, defined appropriately. My idea was to define discretized vectors in such a way that $\langle u^{(K)}, v^{(K)} \rangle$ is the $K$-th Riemann sum for the integral that I would get for $\langle u, v \rangle$ (say, by equally partitioning $[a, b]$ and taking the points of this partition to represent a basis for the finite spaces).

But I know that I need a Lebesgue limit rather than a Riemann limit. Is such discretization possible?

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2 Answers 2

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I think this is always possible when you have a countable orthonormal basis for the Hilbert space. Consider for such a base $\{ e_k \}_{k=1}^\infty$, the spaces $H_K:= \text{span}\big( \{e_k\}_{k=1}^K \big)$.

Then $u^{(k)}=\sum_{k=1}^K \langle u , e_k \rangle e_k$ and $v^{(k)}=\sum_{k=1}^K \langle v , e_k \rangle e_k$. Use the triangle inequality and Cauchy-Schwartz to get that

$$ |\langle u, v \rangle - \langle u^{(K)}, v^{(K)} \rangle| \leq | \langle u-u^{(K)}, v \rangle|+ | \langle u^{(k)}, v -v^{(K)} \rangle| \leq \Vert v\Vert \cdot \| u -u^{(K)} \| + \Vert u^{(K)}\Vert \cdot \| v -v^{(K)} \|. $$

You can now use Bessel's inequality and\or Parseval's inequality to get that $\| u^{(K)} \| \leq \Vert u\|$, while $\| u-u^{(K)} \|^2 \leq \sum_{k=K+1}^\infty | \langle u,e_k\rangle |^2 \overset{K\to \infty}{\to}0$ and likewise for $v-v^{(K)}$.

Now choose $\{ e_k \}$ to be your prefered orthonormal base of $L^2[a,b]$.

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  • $\begingroup$ Thank you, very clear and general. This requires the Hilbert space to be separable, is that correct? $\endgroup$
    – NYG
    Feb 22 at 11:02
  • $\begingroup$ @NYG Yes, otherwise you can't take a sequence of finite dimensional spaces. But the $L^2$ spaces are considering are separable. $\endgroup$ Feb 22 at 14:38
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Keen-ameteur’s answer provides a nice abstract version of this, applicable to all Hilbert spaces. In case of $L^2[a, b]$ though, the following might be closer to what the OP had in mind:

Let $H_K = \mathbb{C}^K$ equipped with the following renormalized inner product:

$$\langle (x_i), (y_i) \rangle = \frac{b-a}{K}\sum_{i=1}^K \bar{x_i}y_i$$

For each $u \in L^2[a, b]$, we then let $u^{(K)} \in H_K = \mathbb{C}^K$ be given by,

$$[u^{(K)}]_i = \frac{K}{b-a} \int_{a+\frac{(i-1)(b-a)}{K}}^{a+\frac{i(b-a)}{K}} u(t) \, dt$$

That is, we are averaging over a partition with width $\frac{b-a}{K}$. One can show that this “discretization” satisfies the desired property.


I’m not sure what is the easiest way to prove $\langle u^{(K)}, v^{(K)} \rangle_{H_K} \to \langle u, v \rangle_{L^2[a, b]}$, but the way I would do this is by first proving the above holds for continuous functions $u$ and $v$. Using uniform continuity, we see that for large enough $K$, $u(t)$ is $\epsilon$-close to $[u^{(K)}]_i$ whenever $t \in [a+\frac{(i-1)(b-a)}{K}, a+\frac{i(b-a)}{K}]$. The same holds for $v$. Then a careful analysis using this fact and the fact that a continuous function must be bounded on $[a, b]$ proves the result for continuous functions. Using Jensen’s inequality, we then see that the map $L^2[a, b] \ni u \mapsto u^{(K)} \in H_K$ is a bounded linear map of norm $1$, for all $K$, so a standard $3\epsilon$-style argument would extend the result to all elements of $L^2[a, b]$, using the fact that continuous functions are dense in $L^2[a, b]$.

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