# A class function $f$ is a character if and only if $(f,\chi_{q_i})_G$ is a non-negative integer, for all irreducible characters $\chi_{q_i}$

I'm currently revising representation theory and I'm a bit stuck trying to prove the converse of the above statement.

$(\Rightarrow)$ is straight forward because if $f$ is the character of a representation $\rho$, then it is a direct sum of irreducible representations $q_i$. And $\rho\sim n_1q_1\oplus...\oplus n_kq_k$ then $(f,\chi_{q_i})_G =n_i$ which is a non-negative integer.

For $(\Leftarrow)$ I'm not sure how to do it.

I know that $\chi_{q_i}$ form a basis for the space of class functions. So $f=\sum_{i=0}^{k} c_i\chi_{q_i}$ where $c_i\in \Bbb{C}$

If $(f,\chi_{q_i})_G =n_i \in \Bbb{Z}_{\ge 0}$ then im guessing the $n_i=c_i$ (not sure why using the definition of $(\cdot,\cdot)_G$).

And maybe $\rho$ will be the direct num of $n_i$ copies on each $q_i$ but we haven't actually shown that $f$ is a character yet?

Any help would be appreciated.

## 1 Answer

The thing to remember is that the $\chi_{q_i}$s are orthonormal with respect to this inner product. In particular if $f=\sum_{i=0}^{k} c_i\chi_{q_i}$ then $(f,\chi_{q_i})_G = c_i$, so by assumption all the $c_i$s are non-negative integers. And indeed we get that $f$ is the character of $\rho\sim c_1q_1\oplus...\oplus c_kq_k$.