# Show that a norm on an inner product space satisfies parallelogram law.

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.

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

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