Do there exist inner product spaces for families of real valued functions other than weighted integrals?

Question

In transform theory we join linear algebra with analysis by defining scalar products for real valued functions with weighted integrals of products, for example:

$$\langle f,g\rangle_w = \int_{-\infty}^{\infty} w(t)f(t)g(t)dt$$

And in multiple dimensions as multivariate integrals:

$$\langle f,g\rangle_w = \int\cdots\int_{-\infty}^{\infty} w(t_1,\cdots,t_k)f(t_1,\cdots,t_k)g(t_1,\cdots,t_k)dt_1 \cdots dt_k$$

To my question, does there exist other ways to define inner products for families of real valued functions?

Answer 1

In terms of interesting and useful examples, I think the $L^2$ Sobolev norms are important: one approach to the definition/characterization of $H^k[a,b]$ for interval $[a,b]$ (for example) with non-negative integer $k$, is as the completion of $C^\infty[a,b]$ with respect to the $H^k$ norm (-squared) $$ |f|^2_{H^k} \;=\; |f|^2_{L^2} + |f'|^2_{L^2} + \ldots + |f^{(k)}|^2_{L^2} $$ The opposite definition/characterization is as distributions $f$ whose first $k$ derivatives are in $L^2$. It is obviously important to understand that these two characterizations give the same thing. :)

And these are not just definitions for the sake of definitions, but very useful ideas in solving differential equations, and other applications (even in number theory, in my own experience).

As @Mars reasonably commented/inquired: how to see that the $H^k$ norms are not topologically equivalent to some cleverly weighted $L^2$ space? Not responding quite the most directly, but perhaps more conceptually: the Sobolev imbedding/inequality proves that (e.g., on a one-dimensional interval) $H^1$ is contained in $C^o$. We can prove that no weighted $L^2$ space is contained in $C^o$... (though, yes, this is not an entirely typical exercise...)

Answer 2

As pointed out, it depends what you mean by "other" and by "integral". One example that comes up in multivariate polynomial approximation is on the space $H = \mathbb{R}[x_1, x_2, \dots x_n]$ equipped with the inner product $$\langle p, q\rangle = \bigg(p(\frac{\partial}{\partial x_1}, \dots \ , \frac{\partial}{\partial x_n} )q \bigg)(0)$$

So apply the $p$ differential operator to $q$ and then evaluate at $0$. An example in two variables is $$\langle xy, x^2 + xy\rangle = \bigg( \frac{\partial}{\partial x} \frac{\partial}{\partial y} (x^2 + xy)\bigg)\bigg|_{(0,0)} = 1$$

This then just needs to be completed to a Hilbert Space. Upon first glance, it seems like this could be very different from any of the usual Hilbert Spaces one typically encounters. However, the Hilbert completion of this is the space of power series in $(x_1, \dots , x_n)$ with square summable coefficients. As a Hilbert Space, it's isomorphic to $l^2(\mathbb{N})$.

Answer 3

If you don't specify real valued functions further, there is also a really strange example: $l^2(\mathbb{R})$, the space of functions $f:\mathbb{R} \to \mathbb{R}$ for which $\sum_{x\in\mathbb{R}} |f(x)|^2 < \infty$ with the scalar product

$$\langle f,g\rangle_{l^2(\mathbb{R})} := \sum_{x\in\mathbb{R}} f(x)g(x)$$

is a well defined Hilbert-space. (Neither the sums nor the small $l$ are a typo and you have to think a bit about well-definedness)

In particular the space only consists of functions $f$ such that $\{x\in \mathbb{R}: f(x) \neq 0\}$ is countable, which means that $f=0$ almost everywhere. It is also not separable and as a result not isomorphic to $l^2(\mathbb{N})$ and thus to any of the other examples (which are all isomorphic to each other).