About a discrete Fourier property

Question

Let $F(G)$ denote the space of functions from the abelian group $G$ to $\mathbb{C}$ and let $f\in F(G)$. Define the discrete furier transform of $f$ as $\hat{f}:\widehat{G}\to \mathbb{C}$ (where $\widehat{G}$ is the set of characters of $G$, ie, group homomorphism from $G$ to $\mathbb{C}^{\ast}$ as

$$\hat{f}(\chi)=\sum_{a\in G}f(a)\bar\chi(a)$$

for some $\chi\in \widehat{G}$

There is a lot of properties of the discrete Fourier transform, which can be proved by definition. One of them is for example the following. For a function, $f\in F(G)$, define $f^{-}\in F(G)$ as $f^{-}(g)=f(-g)$ and similarly with $\hat{f}^{-}:\widehat{G}\to \mathbb{C}$ as $\hat{f}^{-}(\chi)=\hat{f}(-\chi)$ show that

$$\hat{f}^{-}=\hat{f^{-}}$$

The property seems easy to prove via definition. But unfortunately I can get the proof. My attempt is the next.

From definition of the RHS we have $$\hat{f^{-}}(\chi)=\sum_{a\in G}f^{-}(a)\bar\chi(a)$$ If we note that $\bar\chi(a)=\chi(-a)$ (here $\bar \chi(a)$ denotes the complex conjugate of $\chi(a)$), then

$$\hat{f^{-}}(\chi)=\sum_{a\in G}f^{-}(a)\chi(-a)$$ and by definition of $f^{-}$

$$\hat{f^{-}}(\chi)=\sum_{a\in G}f(-a)\chi(-a) $$

Indexing the sum by $-a$ instead of $a$ (since $G$ is a group). Then

$$\hat{f^{-}}(\chi)=\sum_{a\in G}f(a)\chi(a) $$

And from here for get the LHS I should have $-\bar\chi(a)$ instead of $\chi(a)$.

From this I can conclude that $-\bar\chi(a)= \chi(a)$ is true only when $\chi(a)$ is a complex number with real part equal to 0. But this implies that the assertion is not true (and since I obtain the problem from a book, I assume that the problem is actually true).

Any hint, suggestion or counterexample is welcome (I have several days in the problem and I do not get it their solution).

Answer 1

(Basically repeating another answer.) Work through it again. I'd use multiplicative notation for $G$ and $\widehat{G}$ (especially for $\widehat{G}$). I'll also use a subscript for the "minus." Facts that are useful:

• $\chi(a^{-1})=\chi(a)^{-1}=\overline{\chi(a)}$ (homomophism into $S^1$, $\bar{\chi}=\chi^{-1}$),
• $\sum_{h\in H}\phi(h)=\sum_{h\in H}\phi(h^{-1})$ (i.e. indexing by an element or its inverse covers the whole group),
• inversion is an involution $(a^{-1})^{-1}=a$, complex conjugation is also an involution (inversion on the circle).

We have $$ \begin{align*} \widehat{f_-}(\chi)&=\sum_{a\in G}f_-(a)\overline{\chi(a)}\\ &=\sum_{a\in G}f(a^{-1})\chi(a^{-1})\\ &=\sum_{a\in G}f(a)\chi(a)\\ &=\sum_{a\in G}f(a)\overline{\bar{\chi}(a)}=\hat{f}(\bar{\chi})=(\hat{f})_-(\chi). \end{align*} $$

Answer 2

From the definitions for each $\chi\in\widehat{G}$ we have $$\widehat{f^{-}}(\chi)=\sum_{a\in G}f^{-}(a)\overline{\chi(a)}=\sum_{a\in G}f(-a)\overline{\chi(a)}=$$ $$\sum_{a\in G}f(a)\overline{\chi(-a)}=$$ $$ \sum_{a\in G}f(a)\overline{(-\chi(a))}= \hat{f}(-\chi)= \hat{f}^-(\chi).$$