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 $$|x|$$?

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.

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

$$|x|$$ 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.

$$|x-y|$$ 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.

• 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||. – Subhrajyoti Nayak Aug 13 at 14:23
• 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 – Subhrajyoti Nayak Aug 13 at 14:26
• @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. – postmortes Aug 13 at 16:07
• $\|x-y\| := |x|+|y|$ does not make sense. The taxi-cab norm on $\mathbb R^n$ is $\|(x_1,\dots,x_n)\| = \sum_{i=1}^n |x_i|$. – Paul Frost Aug 13 at 23:01
• @PaulFrost oops, good catch! Thank-you :) – postmortes Aug 14 at 5:40