TITLE: Functional equation of irreducible characters
QUESTION [10 upvotes]: I am preparing to an exam in representations of finite groups. I am trying to tackle a problem regarding a characterization of irreducible characters:
Let $f$ be a complex-valued function on a finite group $G$. Then:
$$\forall x,y\in G: f(x)f(y) = \frac{f(1)}{|G|}\sum_{z\in G} f(yzxz^{-1}) \leftrightarrow f \text{ is proportional to an irreducible character}$$
I have made some progress in $\rightarrow$: if $f$ is identically 0, then RHS is implied. Otherwise, it is easy to see that $f$ is a class function (choose $y$ such that $f$ doesn't vanish on it, and notice that $\{yzxz^{-1} | z \in G \}$ is invariant under conjugation of $x$. I thought about proving $\leftarrow$ and then writing $f$ as a combination of irreducible characters. 
I also noticed the following: $f(1)$ is the degree of the representation when $f$ is a character (and it appears often in orthogonality relations). Also, the projection on an isotypical component (corresponding to an irreducible representation $\pi$) of a representation $\rho$ is $\frac{\chi(1)}{|G|}\sum_{g} \overline{\chi(g)}\rho(g)$ (where $\chi$ is $\pi$'s irreducible character).
To sum it all: I have some directions, but haven't used orthogonality relations explicitly, for example. Can you help?
EDIT: Some more advances: I've noticed that if $\phi$ is a class function, the equation is equivalent to 
$$\sum_{z\in G} \phi(x) \phi(z^{-1} y z) \sim \sum_{z \in G} \phi(x (z^{-1}yz))$$
which looks nicer ($\sim$ implies that LHS is proportional to RHS). By putting $y=1$ it is easy to deduce that $\sum_{z\in G} \phi(x) \phi(z^{-1} y z) = \phi(1) \sum_{z \in G} \phi(x (z^{-1}yz))$ (i.e., $\phi(1)$ is the ratio).
Some directions in $\leftarrow$: if $\rho$ is an irreducible representation with a simple character $\chi$, then we can let $\rho ' = \sum_{z \in G} \rho_{z^{-1}yz}$. The equation becomes $Tr(\rho_x)Tr(\rho ') = \chi(1) Tr(\rho_x \rho ')$. The product of characters might suggest use of tensor product.

REPLY [3 votes]: The lecturer provided another proof - slightly different from joriki proof. It is less intuitive and it uses convolution. As it happens in representation theory, many problems might eventually be equivalent and just phrased differently, but it is still cool, and avoids a direct use of Schur's lemma:
If $f(1)=0$, it follows that $f \equiv 0$, and $0$ is the zero multiple of the trivial character.
From now on we'll assume $f(1) \neq 0$. As was observed in my first comment, by putting $y=1$ we get that $f$ is a class function, so we can write $f = \sum_{i=1}^{n} a_i \chi_i$ ($\chi_i$ - the simple characters that span the class functions, $a_i$ - complex coefficients).
Using $\chi_i * \chi_j = \frac{|G|}{\chi_j(1)} \chi _j \delta_{ij}$, we get:
$$f * \chi_j = \frac{a_j |G|}{\chi_j(1)}\chi_j$$

Lemma: if $a_j \neq 0$, $f * \chi_j$ is proportional to $f$.

Since $f \neq 0$, $a_j \neq 0$ for some $j$.
By the lemma (which we'll soon prove) : $\frac{a_j |G|}{\chi_j(1)}\chi_j$ is proportional to $f$. It means that $f$ is proportional to $\chi_j$ (and the ratio is $a_j$), QED.

Proof of lemma: we extend $f$ linearly to $\mathbb{C}G$, i.e.: $f(\sum_{g\in G} b_g g) = \sum_{g \in G} b_g f(g)$. We assume $f(1) = 1$ WLOG. The equation implies:
$$f(y)f(x) = f(yx) \forall y \in \mathbb{C}G, x \in Z(\mathbb{C}G)$$
since the center of group algebra is spanned by the indicators of conjugacy classes.

Put $x = \sum_{g \in G} \chi_j(g^{-1}) g \in Z(\mathbb{C}G)$ (as a linear combination of class functions).
$$\forall y \in G: (f * \chi_j)(y) = \sum_{g \in G} f(yg)\chi_j(g^{-1}) = f(y \sum_{g\in G} \chi_j(g^{-1})g) = f(yx) $$
$$=f(y)f(x)=f(y)\sum_{g\in G} \chi_j(g^{-1})f(g) = |G| \langle f, \chi_j \rangle f(y) = |G| a_j f(y)$$