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In the opening chapter of a functional analysis book I had this question:

Prove that for a norm $||\cdot||$ that if for all vectors $u$ and $v$ it is true that $2||u||^2 + 2||v||^2 = ||u+v||^2 + ||u-v||^2$ then there is an inner product that results in this norm. (That is $\left<u,u\right> = ||u||^2$ ).

So, I figured that if there is an inner product it ought to be definable by $\left<u,v\right> = \frac{||u+v||^2 - ||u-v||^2}{4}$ and I would proceed to show that this satisfies the conditions of inner products.

However, when I go to prove that $\left<\lambda u, v\right> = \lambda \left<u,v\right>$ I'm completely stuck. I can't see any way of demonstrating this.

Does anyone know how I can complete this proof or of an alternate method of proving this?

Rioghasarig
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2 Answers2

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Here are some hints.

The inner product is definable by means of the polarization identity, $$ \langle x,y \rangle = \frac{1}{4}\left(||x+y||^2 - ||x-y||^2\right) + \frac{\rm i}{4}\left(||x+{\rm i}y||^2 - ||x-{\rm i}y||^2\right) . $$ Showing strict positivity & symmetry is easy. Showing linearity under addition is half a page, but very doable (use the parallelogram law).

Showing linearity under scalar multiplication calls for a far deeper insight - at least the way I did it; maybe there are easier ones. Essentially, you have to build your way up from showing such linearity over $\mathbb{Q}_+$ to showing it for $\mathbb{R}_+$ (closure - use continuity), then to $\mathbb{R}$ (enough to see what happens under multiplication by $-1$) & then to $\mathbb{C}$ (split into real & imaginary parts; see what happens under multiplication by $\rm i$). So why is linearity over $\mathbb{Q}_+$ doable? Well, because it boils down to using additive linearity which you showed in half a page above :).

NB: I'd be very interested to know whether someone knows of a simpler proof.

automaton 3
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  • Thank you. But it's strange that this would be so complicated. This is the first chapter and all the other problems could be solved in under a minute. There was no indication given that this problem was particularly challenging. – Rioghasarig Sep 19 '13 at 20:39
  • As far as I know, there isn't really an easier way to prove this. Here are some of those step spelled out a bit more explicitly: http://math.stackexchange.com/questions/21792/norms-induced-by-inner-products-and-the-parallelogram-law – BaronVT Sep 19 '13 at 20:40
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    Thank you, BaronVT. It did take me a couple of days to figure it out back in the day - I now see the older thread appears in the Related column in this page :)... @Rioghasarig - not sure what to say; either the authors have a smarter solution or they pulled a Fermat on their readers. – automaton 3 Sep 19 '13 at 21:47
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Hint: Consider the case that $\langle u, u \rangle = ||u||^{2}$ and expand out $||u+v||^{2} + ||u-v||^{2}$ to see what you get

Keeran Brabazon
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    I don't see how this helps.

    It would help if I was doing the proof the other way around (that is an inner product results in the parallelogram inequality), but I don't think that fact is useful for the converse.

     – Rioghasarig Sep 19 '13 at 20:38
  • This assumes that such an inner product exists, but the question is about the converse. –  Sep 19 '13 at 20:55