Approximating norm of a Hilbert space point with the norm of a vector

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?

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 $$L^2[a,b]$$.

• Thank you, very clear and general. This requires the Hilbert space to be separable, is that correct?
– NYG
Feb 22 at 11:02
• @NYG Yes, otherwise you can't take a sequence of finite dimensional spaces. But the $L^2$ spaces are considering are separable. Feb 22 at 14:38

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 $$L^2[a, b]$$.