Distance of functions and fourier coefficients

1. Nov 21, 2007

sridhar10chitta

There are two functions f(t) and g(t); t is the independent variable.
The distance between the two functions will be given by [1/2pi integral{f(t)-g(t)}^2 dt]^1/2 between -pi and +pi.
Apparently, this distance also is the fourier coefficient of each term in the fourier
expansion of a periodic function f(t) such that it is closest to f(t).

Why is this so ?
why is not the distance given by f(t)-g(t) simply ?
i.e 1/sqrt(2pi) integral{f(t)-g(t)}dt between -pi and +pi.

2. Nov 21, 2007

HallsofIvy

Staff Emeritus
No! The distance between two functions is NOT given by that. It is, rather, given by the square root of that function. The way you have it, the triangle inequality would not be true.
In $L_2$, the space of square integrable functions, it can be show that the finite Fourier series is essentially the "least squares approximation" to the function using that distance (with the square root).

That wouldn't make much sense would it? For one thing, the "distance from f to g" would be the negative of the "distance from g to f"!

You CAN define "distance from f to g" to be $\int |f(t)- g(t)|dt$ where the integral is taken over whatever interval you are interested in. That is the "L1 norm. You can also define "distance from f to g" to be max|f(t)-g(t)| where the "max" is the largest value that takes on over whatever interval you are interested in. That's called the "uniform norm" because convergence defined in that norm is equivalent to uniform convergence. We use the "root square" distance formula in Fourier Analysis precisely because the finite Fourier series is the "least squares approximation" with that distance formula.