Hilbert Space

Resources

Axler, Sheldon. 2020. Measure, Integration & Real Analysis. Vol. 282. Graduate Texts in Mathematics. Cham: Springer International Publishing. https://doi.org/10.1007/978-3-030-33143-6.

Main Idea

More specifically, this is a Vector Space, where the inner product induces a metric such that this Metric Space is complete. This allows us to extend the ideas of Linear Algebra and Calculus to an infinite-dimensional case. A Hilbert space is also a Banach Space. Alternatively, a Hilbert Space is an inner product space and also a Banach Space, where the inner product induces the associated norm. A Sobolev Space is a Hilbert Space.

Examples

A closed subspace of a Hilbert space is also a Hilbert space.
$ℓ^{2}$ with inner product $⟨ (a_{1}, a_{2}, . . .), (b_{1}, b_{2}, . . .) ⟩ = \sum_{k = 1}^{\infty} a_{k} {\overset{―}{b}}_{k}$

Counterexamples

The set of continuous functions on the interval $[0, 1]$ , $C ([0, 1])$ with the inner product $⟨ f, g ⟩ = \int_{0}^{1} f \overset{―}{g}$ . The induced norm is not complete on $C ([0, 1])$ . Consider a piecewise linear function with some slope given by $n$ , over $1 / n$ . Then taking $n \to \infty$ gives a function not in $C ([0, 1])$ .

Details

Distance from a point to a set
Let $V$ be a Normed Vector Space, and a subset $U \neq \emptyset$ , and $f \in V$ . Then, the distance from this element $f$ to the set $U$ is given as

distance (f, U) = inf_{g \in U} | | f - g | | .

Informally, this is the distance (as defined by the norm), to the closest element of $U$ , or potentially limit points of $U$ (which may not be in $U$ ) (uniqueness is also not guaranteed.

However, for a Hilbert Space, the distance is attained by a unique element of $U$ , if $U$ is convex. That is, $\exists! g \in U : | | f - g | | = distance (f, U)$ . Further, in this case, we can define the orthogonal projection as $P_{U} (f) = g$ . Further, if $U$ is a closed subspace (not necessarily convex), then

$f - P_{U} (f)$ is orthogonal to $g$ , $\forall g \in U$ (if it is a projection, it is orthogonal).
If $h \in U$ and $f - h$ is orthogonal to $g$ , $\forall g \in U$ , then $h = P_{U} f$ (if orthogonal and in set, then it is a projection).
$P_{U} : V \to V$ is a Linear Operator (note that we need $U$ to be a closed subspace, making it "linear").
$| | P_{U} (f) | | < | | f | |$ for $f \notin U$ , and $| | P_{U} (f) | | = | | f | |$ if and only if $f \in U$ .