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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 $|x|$?

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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.

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  • $\begingroup$ What if x€R^n and y€R^n ? How do you define ||x-y|| and |x-y| $\endgroup$ – Subhrajyoti Nayak Aug 13 at 13:53
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$|x|$ 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.

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$|x-y|$ 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.

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  • $\begingroup$ Sir essentially norm on a vector space X is a real valued function on X whose value at an x€X is denoted by ||x||. $\endgroup$ – Subhrajyoti Nayak Aug 13 at 14:23
  • $\begingroup$ But some of the books write norm as length of a vector. The way u defined the taxi cab norm ,it does not give the length of the vector x-y. As the taxi cab satisfies all the properties of the norm hence we say it is a norm on Rn $\endgroup$ – Subhrajyoti Nayak Aug 13 at 14:26
  • $\begingroup$ @SubhrajyotiNayak It does give the length of the vector, but it doesn't give the Euclidean length. If you consider just ${\mathbb R}^2$ and draw this on squared paper, then you'll see that this norm is what you get if you can only move parallel to one of the axes. That can be a real restriction: if you are studying the flow of fluid through a rectangular array of pipes, for example. $\endgroup$ – postmortes Aug 13 at 16:07
  • $\begingroup$ $\|x-y\| := |x|+|y|$ does not make sense. The taxi-cab norm on $\mathbb R^n$ is $\|(x_1,\dots,x_n)\| = \sum_{i=1}^n |x_i|$. $\endgroup$ – Paul Frost Aug 13 at 23:01
  • $\begingroup$ @PaulFrost oops, good catch! Thank-you :) $\endgroup$ – postmortes Aug 14 at 5:40

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