Showing inner product of integral function

Question

I'm trying to solve this problem but the textbook has very airy explanations regarding how to do so. I think I can do the first part by just checking off that the 4 axioms of inner products hold.

I'm unsure what the second part is even asking.

Let $L^2((0,1))$ be the space of integrable functions such that $\int_0^1f(t)^2dt<\infty$. Show that $<f|g> = \int_0^1f(t)g(t)dt$ defines an inner product on $L^2((0,1))$.

Show that $B={sin(2{\pi}nt)}_{n\epsilon N}$ is a family of orthogonal vectors with respect to this scalar product.

Any help is appreciated. Thanks.

Answer 1

Two vectors, u and v, are "orthogonal" if and only if = 0. A family of functions is "orthogonal" if and only if any pair of them is orthogonal.

To prove that $sin(2\pi nt)$, where n can be any positive integer, is an orthogonal family you need to show that $\int_0^1 sin(2\pi nt)sin(2\pi mt) dt= 0$ where m and n are positive integers, $m\ne n$.

Answer 2

So, you have a family of vectors $B = \{f_i \}_{i = 1}^{\infty}$.

When do you say that 2 vectors $\overline{a_1}, \ \overline{a_2}$ in simple Euclidean space (let's say $\mathbb{R}^3$) are orthogonal? If it's inner product in this space is equal to zero.

So try to show that inner product of 2 different vectors of set $B$ (vectors in $L^2(0, 1)$ are functions, inner product of 2 functions there is an integral above) is orthogonal.

Answer 3

You need to start from the properties of inner product:

1. $<u+v,w>=<u,w>+<v,w>$
2. $<\alpha u,w>=\alpha<u,w>$
3. $<u,v>=<v,u>$
4. $<v,v>\ge 0$ and equality is achieved only for $v=0$

It should not be difficult to just plug in your definition of the inner product, to see that it verifies these conditions

For part 2, you need to prove that the inner product $<\sin(2\pi nt),\sin(2\pi mt)>=0$ for $n\ne m$ and is not $0$ for any $n=m\ne 0$

Answer 4

We need to prove that $\langle\sin(2\pi mt), \sin(2\pi nt)\rangle=0$ i.e. $$\int_0^1 \sin(2\pi nt)\sin(2\pi mt)dt=0,$$ for $m,n \in \mathbb{N}$, $m \neq n$.

Computing the indefinite integral gives us $$\int \sin(2\pi nt)\sin(2\pi mt)dt=\frac{\frac{\sin(2\pi t(m-n))}{m-n}-\frac{\sin(2\pi t(m+n))}{m+n}}{4\pi}+C, \: C \in \mathbb{R},$$ so $$\int_0^1 \sin(2\pi nt)\sin(2\pi mt)dt=\frac{1}{4\pi}\left[ \frac{sin(2\pi(m-n))}{m-n}-\frac{\sin(2\pi(m+n))}{m+n}\right]=\frac{1}{4\pi}(0+0)=0.$$ Remeber that $\sin(2\pi k)=0$, for every $k \in \mathbb{Z}$.