Consider the inner product: $\langle f,g\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)dx$

Question

Consider the inner product: $\langle f,g\rangle = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)g(x)dx$

Show that $\sin{(x)}$ and $\cos{(x)}$ are orthonormal vectors with respect to this inner product. Then compute $\|\sin{(x)} + \cos{(x)}\|$ induced by inner product.

Please help verify my result.

$\frac{1}{\pi}\int_{-\pi}^{\pi}\sin{(x)}\cos{(x)}dx$ after u-sub I got

$\frac{1}{\pi}\big[\frac{1}{2}\sin^2{(x)}\big]_{-\pi}^{\pi} = 0$

So this result indicates $\sin{(x)}$ and $\cos{(x)}$ are orthonormal because it equals zero.

Then for norm $\|\sin{(x)} + cos{(x)}\|$

$\Rightarrow \frac{1}{\pi}\int_{-\pi}^{\pi}\sqrt{(\sin{(x)}+\cos{(x)})^2}dx = \big[-\cos{(x)} + \sin{(x)}\big]_{-\pi}^{\pi}=0$

Not sure if I just needed to do integrals for this problem.

Answer 1

I think you placed the root sign incorrectly,

$||sin(x)+cos(x)|| = $ $\sqrt{<\sin(x) + \cos(x),\sin(x) + \cos(x)>} = \sqrt{ \frac{1}{\pi}\int_{-\pi}^{\pi}(\sin{(x)}+\cos{(x)})^2 dx} = \sqrt{ \frac{1}{\pi}\int_{-\pi}^{\pi}(1 + 2\sin(x)\cos(x))dx} = \sqrt2$ .

Hope this helps!

Answer 2

$\int\sin x \cos x dx=$ $\int \frac {1}{2}\sin 2x\; dx =\int \frac {1}{4}\sin 2x\; d(2x)=$ $-\frac {1}{4}\cos 2x.$

Therefore $\pi <\sin x,\cos x>=\int_{-\pi}^{\pi}\sin x \cos x\; dx =$ $[-\frac {1}{4}\cos 2x]_{x=-\pi}^{x=\pi}=0.$

$$\text {The norm of $f$ is }\quad \|f\|=\left(\frac {1}{\pi}\int_{-\pi}^{\pi}f(x)^2dx\right)^{1/2}.$$ $$\text { Therefore }\quad \pi \|\sin x+\cos x\|^2=\int_{-\pi}^{\pi}(\sin x+\cos x)^2dx=$$ $$=\int_{-\pi}^{\pi}(2-2\sin x \cos x)dx=\int_{-\pi}^{\pi}2dx-2\pi<\sin x,\cos x>=$$ $$=2\pi-0=2\pi.$$ So the norm is $\sqrt 2. $

For a real normed vector space we have $\|f+g\|^2=\|f\|^2+\|g\|^2+2.$ In two dimensions this is just the Cosine Law for triangles.