# Hilbert space

## In words...

Hilbert spaces are a generalization of the usual Euclidean spaces (such as $\R$ or $\R^2$) to higher and possibly infinite dimensions. This generalization comes with natural extensions of concepts from linear algebra, such as inner products and norms.

Infinite-dimensional spaces are also called function spaces. Indeed, a function can be identified with its values taken at all points of its domain, just as a vector can be identified with the values of its components. If the domain contains an infinite number of points, then there are an infinite number of values defining a function.

Formally, a Hilbert space is a vector space equipped with an inner product inducing a norm with respect to which the space is complete. Completeness means that every converging sequence of elements of the space converges to an element which also belongs to the space.

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## In maths...

### Definition

A Hilbert space $\H$ is a vector space equipped with an inner product $\inner{\cdot}{\cdot}_{\H}$ inducing a norm $$\|x\|_{\H} = \sqrt{\inner{x}{x}_{\H}}$$ with respect to which the space is complete.
Here, completeness means that converging sequences of elements in $\H$ converge in $\H$, i.e., $$\lim_{n\to +\infty} x_n \in\H$$ for any sequence $(x_n)$ of elements in $\H$ satisfying $$\lim_{n\to +\infty} \sup_{p,q > n} \|x_p - x_q\|_{\H} = 0 .$$

### Function spaces

The definition of Hilbert spaces does not impose a finite dimension on $\H$. Thus, we can define infinite-dimensional Hilbert spaces, which we call function spaces. Indeed, a function $f : \X\rightarrow \R$ can be identified with its values $f(\g x)$ taken at all points $\g x\in\X$, just as a vector $\g x$ can be identified with the values $x_j$ of its components. If $\X$ is a subset of $\R^d$, then there is an infinite number of points and values.

Given an input space $\X\subseteq \R^d$, a typical Hilbert space of functions is the space of real-valued and square-integrable functions $L^2(\X)$, which is equipped with the inner product defined by $$\forall f \in L^2(\X),\ g \in L^2(\X),\quad \inner{f}{g} = \int_{\X} f(\g x) g(\g x) \ d\g x .$$ This inner product induces the $L^2$-norm $$\|f\|_{L^2(\X)} = \sqrt{\inner{f}{f}} = \sqrt{\int_{\X} f(\g x)^2 \ d\g x }.$$