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I have a problem with the definition of Hilbert Space and Banach Space.

What is the difference between a Hilbert Space and a Banach Space?

JRR
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    A norm comes from an inner product iff it satisfies the parallelogram identity. Some norms don't (eg L^p for p other than 2), hence the distinction between Banach space (i.e. complete normed space) and hilbert space (i.e. complete inner product). For norms that do satisfy the parallelogram identity, the two notions (Banach and hilbert) are indeed equivalent; you get an inner product from a norm by the polarization formula. – ziggurism Dec 03 '15 at 18:52
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    Do you know both definitions? (If you know the definitions I don't see the problem - the difference is exactly what the definitions say. If you don't know the definitions you have no business asking the question before learning the definitions.) – David C. Ullrich Dec 03 '15 at 18:52
  • Hilbert space = Banach space + $\langle \cdot, \cdot \rangle$ –  Dec 03 '15 at 18:54
  • @AhmedHussein: that seems a tad too simple. Is there an inner product can I add to L^1 to get a Hilbert space? – ziggurism Dec 03 '15 at 19:01
  • @ziggurism what I meant is: "A Banach space which has its norm induced by an inner product". –  Dec 03 '15 at 19:08
  • @AhmedHussein still don't like it. Can't just add an inner product to a Banach space, unless the norm is compatible, in which case the addition is redundant. – ziggurism Dec 03 '15 at 19:11
  • @ziggurism, I'm not "just adding" an inner product to a Banach space; I don't know if that even makes sense. What I meant was what I stated in my last comment; in case I'm unaware of something in this regard, please pinpoint it to me. –  Dec 03 '15 at 19:17
  • See Polarization identity. – Dave L. Renfro Dec 03 '15 at 19:23
  • @AhmedHussein "just add inner product" definitely makes sense; that's actually the definition of inner product space (vector space + inner product). But if that's not what your intended with the + in your original comment, please clarify. – ziggurism Dec 03 '15 at 22:48

2 Answers2

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A Hilbert space is a Banach space. What makes it special is that its norm satisfies a parallelogram law: $$\| x-y \|^2 + \| x + y\|^2 = 2 \|x\|^2 + 2\|y\|^2.$$ When the norm obeys the parallelogram law, there is an inner product $\langle x, y \rangle$ such that $\langle x, x \rangle = \|x\|^2$. That is, the inner product induces the norm.

The specialness of Hilbert spaces comes from the inner product. There it is possible to define angles in infinite dimensional spaces that has some resemblance to our intuition about Euclidean spaces. In particular we say that two vectors are at $90^\circ$ (or they are orthogonal) with one another if they satisfy $\langle x, y \rangle = 0$.

Moreover, not all Banach spaces are reflexive, like Hilbert spaces are. Hilbert spaces are infact more than just reflexive, their dual space is isometrically isomorphic to the Hilbert space itself.

Recall that the dual of a Banach space is the collection of continuous linear functionals. The Riesz representation theorem declares that for every continuous functional $T$ on a Hilbert space, $H$, there is a vector $y \in H$ such that $T(x) = \langle x, y \rangle$ and $\| T \| = \|y\|_H$. Thus $T$ can be identified with $y$. If we let $H'$ be the dual of $H$ we then have $H=H' = H''$, and the identity $H=H''$ is called the reflexive property.

If you consider the Banach space $c_0(\mathbb{N})$, the dual space of $c_0(\mathbb{N})$ is $l^1(\mathbb{N})$, and the dual space of $l^1(\mathbb{N})$ is $l^\infty(\mathbb{N})$. Thus $c_0(\mathbb{N})$ is not reflexive. This also demonstrates that the norm for $c_0(\mathbb{N})$ does not arise from an inner product, since otherwise the space would be reflexive. The same can be said of $l^1(\mathbb{N})$, since it is not self-dual.

Definitions:

$c_0(\mathbb{N}) = \{ (a_n)_{n\in \mathbb{N}} \subset \mathbb{C} : a_n \to 0 \}$ and has the norm $\|(a_n)\| = \sup_{n} |a_n|$.

$l^1(\mathbb{N}) = \{ (a_n)_{n\in \mathbb{N}} \subset \mathbb{C} : \|(a_n)\| = \sum |a_n| < \infty \}$.

$l^\infty(\mathbb{N}) = \{ (a_n)_{n\in \mathbb{N}} \subset \mathbb{C} : \| (a_n) \| = \sup_{n} |a_n| < \infty \}$.

Joel
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A Hilbert space is a particular type of Banach space, which has an inner product that induces the norm, via $$ \|x\|=\langle x,x\rangle^{1/2}. $$

Martin Argerami
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