Uniqueness Of Fourier Series.

Question

The book said:" if $f$ and $g$ have the same Fourier Coefficients, then $f$&$g$ are necessarily equal." and it said that"by taking the difference $f - g,$ this proposition can be reformulated as: if $\hat{f}(n) = 0$ for all $n \in \mathbb{Z}$, then $f = 0.$"

I do not understand how the second statement is a formulation of the first statement? Could anyone explain it for me?

thanks.

Answer 1

It is because the Fourier series is linear. Namely, if $f,g$ are two functions, then the Fourier coefficients of $h\stackrel{\rm def}{=} f-g$ are $$ \widehat{h}(n) = \widehat{f-g}(n) = \widehat{f}(n)-\widehat{g}(n) $$ for all $n\in\mathbb{Z}$.

So the two statements

(1) For all $h$, $\widehat{h}(n)=0$ for all $n\in\mathbb{Z}$ implies $h=0$.

and

(2) For all $f,g$, $\widehat{f}(n)=\widehat{g}(n)$ for all $n\in\mathbb{Z}$ implies $f=g$.

are equivalent.

The first implies the second by applying (1) to $h\stackrel{\rm def}{=} f-g$; the second implies the first by applying (2) to $f\stackrel{\rm def}{=}h$, $g\stackrel{\rm def}{=}0$.