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- Thread starter Manchot
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HallsofIvy

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|ai+ bj+ ck|= [tex]\sqrt{a^2+ b^2+ c^2}[/tex]

|ai+ bj+ ck|= |a|+ |b|+ |c|

|ai+ bj+ ck|= max(|a|,|b|,|c|).

If you look at the set of vectors that have norm equal to, say, 1, the first (standard) norm gives the surface of a sphere. The second norm gives the surface of "diamond" formed of two pyramids. The third norm gives the surface of cube. All "neighborhoods" of points using those norms are of those forms.

You then define a set to be "open" if around every point in the set you can find a neighborhood of the point that is completely in the set. But it is easy to show that given any one of the different kinds of neighborhoods (different norms) you can find either of the other kinds of neighborhoods

HOWEVER, it is possible to choose unusual norms where that is not true. For example, the "discrete" norm is defined by : "the norm of the zero vector is 0, the norm of any non-zero vector is 1". In that case,

Your use of "L

We also often use "L

And, of course, there is the "uniform norm": If f(x) is a continuous function on a compact set, define the norm to be ||f(x)||= max |f(x)| where the max is taken over the entire set. That is called the "uniform" norm because convergence of a sequence of functions in that norm is uniform convergence.

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Hurkyl

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matt grime

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Manchot said:

The answer is "yes and no" depending on what space you're talking about.

Yes, there are different calculuses we can define. The reals are the clsure of the rationals in the euclidean norm. there are also "valuations" that give the p-adic numbers, one for each prime p. all of these are distinct.

however, if we try to put a norm on any finite dimensional vector space then the resulting topological spaces are all equivalent, that is they define the same topology.

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||a|| = <a,a> ^ (1/2)

Then how could we define our inner product to get those other norms you mentioned?

Is this always possible to do? I mean, can one always find an inner product for any given norm?

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mathwonk

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there is a notion of "equivalence" of norms, where two norms, ||x|| and |x|, are equivalent iff there exist positive numbers c,d such that for all x: |x| < c||x||, and ||x|| < d|x|. then equivalent norms define the same notions of convergence and of derivatives.

if you compute these constans for R^n, and then let n go to infinjity, you can see why the three norms halls gave, become inequivalent in infinite dimensions.

halls is thinking of a "discrete metric", which does not however derive from a norm.

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mathwonk

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also not all norms derive from inner products, except in finite dimensions.

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matt grime

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the cirterion for getting an inner product from a norm is called the polarization identity. i can't recall it exactly off the top of my head, so try googling for something on the wolfram website.

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Wolfram doesn't have it, but wikipedia says

[tex]

\[

\left\| {x + y} \right\|^2 = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >

\]

[/tex]

is satisfied in a vector space whose norm is defined in terms of an inner product.

Thats neat, I had discovered this identity for myself before and thought it was interesting since it was just like expanding a binomial.

[tex]

\[

\left( {x + y} \right)^2 = x^2 + y^2 + 2xy

\]

[/tex]

I had no idea it had a name though.

(edit: fixed the typo matt grime mentioned)

[tex]

\[

\left\| {x + y} \right\|^2 = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >

\]

[/tex]

is satisfied in a vector space whose norm is defined in terms of an inner product.

Thats neat, I had discovered this identity for myself before and thought it was interesting since it was just like expanding a binomial.

[tex]

\[

\left( {x + y} \right)^2 = x^2 + y^2 + 2xy

\]

[/tex]

I had no idea it had a name though.

(edit: fixed the typo matt grime mentioned)

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matt grime

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shmoe

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matt grime said:

The parallelogram law is:

[tex]2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2[/tex]

The special property you referenced is the fact that if you have a norm, then there exists an inner product such that [itex]\|x\|^2=\langle x,x\rangle[/itex]

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The parallelogram law is always true if the polarization identity is true...

[tex]

\[

\left\| {x + y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >

\]

[/tex]

implies:

[tex]

\[

\left\| {x - y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 - 2 < x,y >

\]

[/tex]

So we can add the two equations and get the parallelogram law.

So, the criteria master coda quoted could just as easily have been: there exists an inner product such that [itex]\|x\|^2=\langle x,x\rangle[/itex]

if and only if the polarization identity is true.

(sorry about the typo in the last post, I fixed it now)

[tex]

\[

\left\| {x + y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >

\]

[/tex]

implies:

[tex]

\[

\left\| {x - y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 - 2 < x,y >

\]

[/tex]

So we can add the two equations and get the parallelogram law.

So, the criteria master coda quoted could just as easily have been: there exists an inner product such that [itex]\|x\|^2=\langle x,x\rangle[/itex]

if and only if the polarization identity is true.

(sorry about the typo in the last post, I fixed it now)

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shmoe

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yeah, I see what you mean, thanks

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mathwonk

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i meant "up to equivalence" of norms.

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