If you have a dot product in 'function space'
$$\langle f,g\rangle$$
This translates to the integral for functions
$$ \langle f,g\rangle =\int f \bar{g} dx $$ If I am not mistaken.
If we have for $g(x)$
$$g(x)=e^{ikx}$$
Then we get $$ \langle f,g\rangle =\int f \cdot e^{-ikx} dx $$
This looks similar to the fourier transform:
$$ f(k)=\frac{1}{\sqrt{2\pi}}\int f(x)\cdot e^{-ikx} dx $$
I think this makes sense because we are calculating how much the function $f(x)$ is pointing in the direction of the base vector (or function) $e_k = e^{ikx}$
I don't understand two things:
- why do you have to take the complex conjugation of $g$?
- Also where does the factor $\frac{1}{\sqrt{2\pi}}$ come from? I suspect it comes from normalizing the base vector, but I don't know clearly.
Could someone help me show how both of these can be understood intuitively?
All inner products want to satisfy the identity $\langle f,f\rangle \ge 0$, which is why we take the complex conjugate. When $f$ is potentially complex, there is no reason why $f^2$ (or its integral) should be nonnegative. On the other hand, $f \bar f$ simplifies to $|f|^2$, which is guaranteed to be nonnegative, and when we integrate it, we get a nonnegative value.
The reason for the factor $\frac{1}{\sqrt{2\pi}}$ is that the function $g_k$ we should be looking at is $g_k(x) = \frac1{\sqrt{2\pi}} e^{ikx}$. (I've added a subscript to $g$ to remind us that there's many such functions.) The reason we look at this function $g_k$ is that it satisfies $$\langle g_k,g_k\rangle = \int_0^{2\pi} g_k(x) \overline {g_k(x)}\,dx = \int_0^{2\pi} \frac{e^{ikx}}{\sqrt{2\pi}} \cdot \frac{e^{-ikx}}{\sqrt{2\pi}} \,dx = \int_0^{2\pi} \frac1{2\pi}\,dx = 1.$$ Together with the fact that $\langle g_k, g_l\rangle = 0$ when $k \ne l$, this makes the sequence $g_0, g_1, g_2, \dots$ an orthonormal basis of a vector space.