From: jhnieto@my-dejanews.com
Subject: Re: Which normed vector spaces are inner product spaces?
Date: Tue, 02 Feb 1999 13:33:08 GMT
Newsgroups: sci.math
To: chanita@flash.net
Keywords: Parallelogram law implies an inner product exists

In article <36B54A5B.E6AD0AB4@flash.net>,
  Chanita Chantaplin <chanita@flash.net> wrote:
> > I have been asked to show that an inner product can be defined on normed
> > spaces which obey the parallelagram law:
> >
> > || f + g ||^2 + || f - g ||^2 = 2 || f ||^2 + 2 || g ||^2
> >
> > It seems to me that the way you would do this is to define a function by
> > the polerization formula (assume for simplicity that the scalar field is
> > the reals):
> >
> > <f,g> = ( || f +g ||^2 - || f-g ||^2 ) / 4
> >
> > and then show that this function satisfies all axioms of an inner
> > product.  The axioms I am having trouble with are
> >
> > < kf, g > = k <f, g>
> > < f+g, h > = < f,h > + < g,h >
> >

First add and subtract ||f-g+h||^2 to 4<f+g,h> and
apply the polar identity to obtain:

  4<f+g,h> = ||f+g+h||^2 - ||f+g-h||^2
= ||f+g+h||^2 - ||f-g+h||^2 + ||f-g+h||^2 - ||f+g-h||^2
= 2(||f+h||^2 + ||g||^2) - 2(||f||^2 + ||g-h||^2)    (1)

Now permute f and g:

  4<g+f,h>
= 2(||g+h||^2 + ||f||^2) - 2(||g||^2 + ||f-h||^2)    (2)

Sum (1) and (2) and divide by 8 to obtain

 <f+g,h> = (1/4)(||f+h||^2 - ||f-h||^2) + (1/4)(||g+h||^2 - ||g-h||^2)
         = <f,h> + <g,h>

From  this identity it follows easily (by induction) that

 <nf,h> = n<f,h> for all positive integers n.

From  the definition of <f,h> we see that <0f,h> = 0 = 0<f,h>
and <-f,h> = -<f,h>, hence <nf,h> = n<f,h> holds for all integers.

if m and n are integers, n>0, then

n<(m/n)f,h> = <n(m/n)f,h> = <mf,h> = m<f,h>

hence <(m/n)f,h> = (m/n)<f,h>

Finally observe (i.e. prove!) that <f,h> depends continuously
of f, hence if q_n is a sequence of reals that converges to a
real number r we have

<rf,h> = lim <q_n f,h> = lim q_n<f,h> = r<f,h>

Greetings,
Jose H. Nieto

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