1
$\begingroup$

Show that a norm on an inner product space satisfies parallelogram law. Hence use the parallelogram law to show that the space of continuous real functions defined on the interval $[a,b]$ is not a Hilbert space. Here I did the first part. Let $X$ be an inner product space and $x,y\in X$ consider $\Vert x+y\Vert ^2$ and $\Vert x-y\Vert ^2$ Then on adding I get $\Vert x+y\Vert ^2+\Vert x-y\Vert ^2=2(\Vert x\Vert ^2 +\Vert y\Vert ^2)$ For the second part I don’t know how can a parallelogram law prove $C[a,b]$ is not a Hilbert space, help me please.

$\endgroup$
  • 1
    $\begingroup$ If $C([a,b])$ were a Hilbert space then the parallelogram law would hold. Show it does not. $\endgroup$ – user20266 May 27 '12 at 11:28
1
$\begingroup$

Consider functions $$ x(t)=\cos\left(\frac{\pi}{2}\frac{t-a}{b-a}\right)\qquad y(t)=\sin\left(\frac{\pi}{2}\frac{t-a}{b-a}\right)\qquad $$ then $$ \Vert x+y\Vert=\sqrt{2},\qquad\Vert x-y\Vert=1\qquad \Vert x\Vert=\Vert y\Vert=1 $$

$\endgroup$
  • $\begingroup$ How did you get $\sqrt{2}$? Should it not be instead $||x+y||={\max}_t(x(t)+y(t))=1+1=2$? $\endgroup$ – Pedro Gomes May 8 '18 at 16:31
  • $\begingroup$ @PedroGomes, please plot the graph of x+y $\endgroup$ – Norbert May 8 '18 at 17:24
  • $\begingroup$ How would I manipulate the expressions to the point of getting $\sqrt{2}$? Thanks for the reply! $\endgroup$ – Pedro Gomes May 8 '18 at 20:31
  • $\begingroup$ @PedroGomes $\sin x +\cos x=\sqrt{2}\sin(x+\frac{\pi}{4})$ $\endgroup$ – Norbert May 8 '18 at 21:17
  • $\begingroup$ Thanks! Finally got it! $\endgroup$ – Pedro Gomes May 9 '18 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.