# Using a different definition of norm

1. Oct 19, 2006

### neutrino

Here's a question from Apostol's Calculus Vol1

Suppose that instead of the usual definition of norm of a vector in $$V_n$$, we define it the following way,

$$||A|| = \sum_{k=1}^{n}|a_k|.$$

Using this definition in $$V_2$$ describe on a figure the set of all points $$(x,y)$$ of norm 1.

Is it possible to do that? Doesn't every point $$(x,y)$$ of the form $$(\frac{1}{s}, \frac{s-1}{s}), s \geq 1$$ satisfy the condition? (i.e., the number of points is not finite)

Last edited: Oct 19, 2006
2. Oct 19, 2006

### matt grime

What has the number of points got to do with the ability to describe them? There are an uncountable number of points of normal norm one, and they are described as the unit circle.

If you just take the point (x,y) and consider the 4 cases of the signs of x and y you will see the answer quite easily (in particular, if x and y are both positive, it is the set { (t,1-t) : 0<=t <=1 }

3. Oct 19, 2006

### neutrino

Thank you. I knew I was forgetting something really simple.

4. Oct 20, 2006

### HallsofIvy

Staff Emeritus
By the way (this may be the next problem in Apostle!), the norm defined by $$||A||= max(|x_k|)$$ is also equivalent to the "usual" norm.

Where the set of all point p such that ||p||= 1, with the usual norm form a "ball" and the set of all points, ||p||= 1 ,with the norm you give, form a "diamond", the set of all points, ||p||= 1, with this norm, form a "cube".

The norms are equivalent since given any one, we can find a smaller of the other figures that will fit inside. For any neighborhood in one norm, there exist neighborhoods in the other norms that fit inside. Thus, open sets are identical in all 3 norms.

This is for finite dimensional space. Interestingly, for infinite dimensional spaces, such as function spaces, the 3 norms:
$$||f||=\sqrt{\int f(x)^2 dx}$$
(The "L2" norm)
$$||f||= \int |f(x)|dx$$
(The "L1" norm)
$$||f||= max |f(x)|$$
(The "$L_\infty$" norm)
do not give the same things. In fact, the sets of functions for which they exist are different.

5. Oct 20, 2006

### neutrino

Indeed it is. :)

That is very interesting! Although it may not be very difficult to do it in V_3, I usually stick to V_2 for visualising these concepts.
Now, that is somewhat beyond what I can understand. I'm self-studying linear algebra (just begun, in fact), in a non-math-methods way limited to computing determinants and solving linear equations. But I'm sure I'll be able to appreciate this in the future. Thanks.