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My question regard the relationship between discrete and continuous inner product $\langle f(x), g(x)\rangle =\int_a^b f(x)\overline{g(x)}dx=\lim_{N\to \infty}\sum_{i=0}^N f(a+(b-a)i/N)\overline{g(a+(b-a)i/N)}\frac{b-a}{N}\\$

$\langle \textbf{x}, \textbf{y}\rangle = \sum_{i=1}^N \textbf{x}(i)\overline{\textbf{y}(i)}$

It is the $\frac{b-a}{N}$ that confuses me: i know the definition of riemann integral and the infinite sum definition (without the dx term the sum would diverge) But with this term we cannot express the continuous inner product as a limiting case of the discrete one.

If the dimension of $\textbf{x}$ is $N$ we cannot simply say that continuous time inner product is the limit case of the discrete one. Any suggestion?

A similar question was already proposed in Understanding dot product of continuous functions but to me it is not clear the $\frac{b-a}{N}$ term

Forgive my poor english and my poor formalism (I am writing form a smartphone and using latex is very difficult), thank to everyone

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ernst
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  • When you take the limit of the finite dimensional one you get a countable dimension (e.g. polynomial) but your function space has an uncountable dimension. You got things backwards : when defining the Lebesgue integral, you can unify both cases by writing the second one as an integral. – xavierm02 Sep 26 '14 at 10:56

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If $f$, $g$ are continuous on $[a,b]$, the Riemann integral can be viewed as a limit of discrete inner products: $$ \int_{a}^{b}f(x)g(x)\,dx = \lim_{\|\mathcal{P}\|\rightarrow 0} \sum_{\mathcal{P}}f(x_{j}^{\star})g(x_{j}^{\star})\Delta_{j}x $$ where $\sum_{\mathcal{P}}\Delta_{j}x=b-a$. So this looks like a weighted inner-product where the weights always add to $b-a$, which allows you to take such a limit. Without the sum of the weights always remaining bounded, the limit would get out of hand.

Disintegrating By Parts
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  • so this was my first interpretation: the continuous function can have an inner product if we use a weight function (in this case as you explained is the $\Delta_{j}x$ factor. This weight function is not present in the discrete case, or at least it is unitary (that is obviously the same thing). So my new question now is:does it exist a general weight function that can fit in both definitions? – ernst Sep 26 '14 at 15:36
  • If $\alpha_{j}$ are positive numbers, then $(x,y)=\sum_{j=1}^{N}\alpha_{j}x_{j}y_{j}$ is an inner-product on $\mathbb{R}^{N}$, for example. If $A$ is a positive definite $N\times N$ matrix, then $(x,y){A}=(Ax,y)$ is also an inner-product, which, after a change of basis, is a weighted inner product with some positive $\alpha{j}$, at least when viewed in the coordinate system of the eigenvectors of $A$. Similarly, if $\rho > 0$, then $\int_{a}^{b}f(x)g(x)\rho(x),dx$ is an inner product. Or, more generally, if $\mu$ is a positive measure on $[a,b]$, then $\int fgd\mu$ is an inner product. – Disintegrating By Parts Sep 26 '14 at 17:21