Showing a orthogonal basis is complete

Question

$\psi_1 = \frac{1}{\sqrt{2}}$

$\psi_2 = \sqrt{\frac{3}{2}}x$

By shwoing that any arbitrary function $f(x)=ax+b$ can be represented as linear combination of $\psi_1$ and $\psi_2$, show that $\psi_1$ and $\psi_2$ constitute a complete basis set for representing such functions

So I showed that $\psi_1$ and $\psi_2$ are orthonormal by taking their inner product. Is that all that I have to do? Because then we know that for any constants, this will still be true, and we can cover any real number or imaginary number by multiplying it in.

I would really appreciate any help, this seems a little too simple, so i feel as tho i am missing something.

Answer 1

Yes, given that you've already done the first step. To prove that something is an orthonormal basis you will have to prove two things:

1. That it is a basis - that is that you can represent all the elements as a linear combination of them.
2. That it is orthonormal - that is that $\langle \psi,\phi\rangle$ is one if $\psi = \phi$ and zero otherwise.

Actually there's one more requirement for the first - to be a basis they should be linearly independent. But it follows from orthonormality, say you have a linear combination $\psi=\sum c_j\psi_j$ with some $c_j\ne 0$ then you have $\langle \psi, \psi \rangle = \sum\sum c_j\overline{c_k} \langle\psi_j,\psi_k\rangle = \sum |c_j|^2\langle\psi_j, \psi_j\rangle= \sum |c_j|^2 \ne 0$ so such a linear combination can't become zero (which means they're linearly independent).