Hilbert Spaces and Banach Spaces 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?
 A: A Hilbert space is a particular type of Banach space, which has an inner product that induces the norm, via
$$
\|x\|=\langle x,x\rangle^{1/2}.
$$
A: 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 $\|(a_n)\| = \sup_{n} |a_n|$.
$l^1(\mathbb{N}) = \{ (a_n)_{n\in \mathbb{N}} \subset \mathbb{C} : \|(a_n)\| = \sum |a_n| < \infty \}$.
$l^\infty(\mathbb{N}) = \{ (a_n)_{n\in \mathbb{N}} \subset \mathbb{C} : \| (a_n) \| = \sup_{n} |a_n| < \infty \}$.
