Are all Banach spaces also Hilbert spaces? We have the well-known "polarization identity"
$$(x,y)=\frac{1}{4}\left(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right)\tag{1}$$
that works in any Hilbert space. Hence, is every Banach space also a Hilbert space by just defining the inner product $(\cdot,\cdot)$ by $(1)$?
The claim is clearly false, but I cannot find why.
 A: In most Banach spaces, the thing defined by (1) is not an inner product; it will not satisfy the axioms. A necessary and sufficient condition for the norm to come from an inner product is that the Parallelogram Law holds, $2\|x\|^2+2\|y\|^2 = \|x+y\|^2+\|x-y\|^2$ for all $x$ and $y$ in the space.  You only need the necessary part for your question, which is easy to check directly from the definition of a norm defined by an inner product.  Thus to show that a given norm doesn't come from an inner product, it suffices to find elements $x$ and $y$ such that the above equation doesn't hold.  
Example: Take $\mathbb C^2$ with $\|(x_1,x_2)\|=\max\{|x_1|,|x_2|\}$.  This norm does not come from an inner product, because with $x=(1,0)$ and $y=(0,1)$, you have $2\|x\|^2+2\|y\|^2=4$ while $\|x+y\|^2+\|x-y\|^2 = 2$.  
A: A Banach space is a complete metric space with a norm, while a Hilbert space is a complete metric space with an inner product. An inner product induces a norm by $||x|| = \sqrt{<x,x>}$, which means that any Hilbert space is also a Banach space. However, the converse need not hold as the norm isn't always expressable in terms of the inner product.
An example is $C(K, \mathbb{R})$, the space of all continuous functions on a compact  set $K$ into $\mathbb{R}$ with the norm $||f|| = \sup_{x \in K}{|f(x)|}$. 
