In Serre's "Linear Representations of Finite Groups" he proves (Theorem 6, p.19) that…
The irreducible characters $\chi_1,\ldots,\chi_r$ of a finite group $G$ span the space of class functions on $G.$
(By this point in the book, he has already proved that these characters form an orthonormal system.)
His proof starts like this:
Suppose that $f:G \to \mathbb{C}$ is a class function which is orthogonal to every $\chi_i^*.$
For each representation $(V,\varrho)$ of $G,$ consider the following $G$-equivariant endomorphism of $V$:
$$\hat{f}\!(\varrho)=\sum_{g \in G} f(g)\varrho(g).$$
He shows that, for an irreducible representation $(V,\varrho),$ we have $\hat{f}\!(\varrho)=0.$
Then, and this is where I lose him, he says:
From the direct sum decomposition we conclude that $\hat{f}\!(\varrho)$ is always zero.
I don't see any kind of additivity relation like $\hat{f}\!(\varrho\oplus \sigma) = \ldots$
Question. How does he conclude $\hat{f}\!(\varrho)=0$ for all $\varrho$ by knowing it for all irreducible $\varrho$?
Best Answer
If $V=V_1\bigoplus V_2\bigoplus\cdots\bigoplus V_n$, with each $V_k$ irreducible, then, for each $v\in V_k$,$$\hat f(\rho)(v)=\sum_{g\in G}f(g)\rho_g(v)\in V_k,$$since $V_k$ is a subrepresentation of $V$. So, $\hat f(\rho)(V_k)\subset V_k$ and so, since $V_k$ is irreducible, $\hat f(\rho)|_{V_k}=0$. Since this occurs for each $V_k$, $\hat f(\rho)=0$.