Normed Linear Spaces I am taking a first course in functional analysis but I am unable to understand the difference between $\|x-y\|$ and $|x-y|$ ?
I have doubt that if $|x-y|$ is defined as the distance between x and y ,then why to we   define $\|x-y\|$  and what do we literally mean by that?
I too have the same problem with $\|x\|$ and    $|x|$?
 A: Well, the notation $|x|$ is the absolute value of a real number (or complex number) $x$, while $\|x\|$ is the norm of a vector $x$.
In particular, if $x$ and $y$ are real-valued vectors of length $n$, then $x-y=(x_1-y_1,\ldots,x_n-y_n)$ is also a vector and the norm definition applies.
A: $|x-y|$ is an example of $\|x-y\|$.  You can have other definitions for $\|x-y\|$ that are different; for example, the taxi-cab norm (also called the Manhatten norm, but properly known as the $\ell_1$ norm.  That norm is $\|x\| := |x_1|+|x_2|+\cdots |x_n|$ for ${\mathbb R}^n$.  This extends as you'd expect: for $x,y \in {\mathbb R}^n$ you have 
$$\|x-y\| = \sum_{i=1}^n |x_i - y_i|$$
Picking up from the comments: a norm defines a length on a space, not necessarily the Euclidean one.  If you have a constraint, for example, that you can only move parallel to co-ordinate axes (think of a 3D printer head which can be moved left-right and forward-backward and up-down but is otherwise fixed).  The $\ell_1$ norm is then the correct norm to use when working out the shortest distance between points.  If you're working in an environment where there is a penalty to be paid for moving in a certain direction then a weighted norm will allow you to calculate the shortest distance that incurs least penalty (for a simple example, using the 3D printer head again, you might want to ensure that everything at a certain height is plotted before moving to the next height so you could modify your norm to be $\|x-y\| = |x_1-y_1| +|x_2 - y_2| + \alpha |x_3-y_3|$ where $\alpha >0$ is the penalty for moving up.
A: $|x|$ is the absolute value of a (real/complex) number. For real numbers it is defined as
$$|\cdot|:ℝ→ℝ_{\ge 0},\qquad x\mapsto \begin{cases}x & x \ge 0\\ -x & x < 0.\end{cases}$$
$||x||$ is the norm of a vector space element $x∈V$. There are many different possible norms on each vector space.
So $|\cdot|$ operates on an ordered field while $||\cdot||$ operates on a vector space. They are quite different things.
But on the other hand real/complex numbers also are a 1-dimensional vector spaces, and one possible norm on these vector spaces is the absolute-value norm $||x||:=|x|$. For them norm and absolute value is more or less the same.
