Parallelogramity
Feb. 1st, 2006 12:07 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png)
So. Got the parallelogram law problem. Basically, a big part of the reason I wasn't getting additivity before was that I was convinced it was much harder than it really was, that you couldn't just use parallelogram law and you needed to somehow do it as two inequalities using triangle inequality. Somehow I recalled Tom saying he did it by triangle inequality; probably I just misremembered. Once Lucas told me he did it using just parallelogram law, I got it within a few tries. (Well, also after he pointed out another way you can use parallelogram law (isolate ||v+w||²), after which I got it immediately (though actually not the way he did it, a more direct way (he went through <v_1+v_2,w>=<2v_1,w>+<v_2-v_1,w>))). So it's actually not such a hard problem after all... I actually came very close a few days ago, but I tried to use triangle inequality instead of parallelogram law and didn't see it!
Hm, so since you actually don't use triangle inequality at all, that means that if ||•|| is positive-definite, scalars come out in absolute value, and satisfies the parallelogram law, then it's automatically a norm generated by an inner product and satisfies triangle inequality as well! I doubt that that's actually much use in anything, but it's neat to note. This is wrong; see below.
-Sniffnoy
-Sniffnoy
Re: Homogeneity?
Date: 2006-02-01 10:31 am (UTC)However, ISTM that you don't need scalars-come-out for the real case, while for the complex case, all you need is ||iv||=||v||, ||(1+i)v||=2||v||.
...oh wait, then you again don't have a norm in the first place for continuity. ...no, but wait, because of triangle inequality, and since parallelogram law implies ||-v||=||v||, you do still get a metric, so I suppose actually it does still work...
Re: Homogeneity?
Date: 2006-02-03 08:43 pm (UTC)