Character Theory Exercise I am having trouble with the following exercise in character theory:
If $\chi, \psi, \zeta$ are irreducible characters of a finite group then $\langle \chi\psi, \zeta \rangle \leq \zeta(1)$.
I can show that $\langle \chi\psi, \zeta \rangle = \dim \textrm{Hom}_G(U, V\otimes_\mathbb{C} W)$ where V, W, U are the irreducible G-modules corresponding to $\chi, \psi, \zeta$ respectively, but it seems that I have merely translated the question into one about homomorphisms and tensor products. It doesn't help me much or at least I can't see how it does. I would appreciate any hints.
 A: Hint: writing $U=\displaystyle\bigoplus_{i=1}^n U_i$ as a sum of $1$-dimensional $\Bbb C$-subspaces, there is an injection
$$\hom_G\left(\left(\bigoplus_{i=1}^n U_i\right)\otimes V^*,W\right)\to\bigoplus_{i=1}^n \hom_G\big(U_i\otimes V^*,W\big)$$
(Thinks of linear maps $U\otimes V^*\to W$ as bilinear maps $U\times V^*\to W$.)
A: This is Problem (4.12) from I.M. Isaacs, Character Theory of Finite Groups and can be proved much more easier. If $\chi \in Irr(G)$, then observe that $|\chi|$ is a (non-negative real-valued) class function, with $[|\chi|,|\chi|]=1$! (It does not have to be a character by the way: example $G=S_3$, if $\lambda$ and $\mu$ are its linear characters and $\chi$ its character of degree $2$, then $|\chi|=\frac{2}{3}\lambda+\frac{2}{3}\mu+\frac{1}{3}\chi$.) We need a lemma.

Lemma Let $\chi, \psi \in Irr(G)$, then $[|\chi|,|\psi|] \leq 1$.

Proof Note that $[|\chi|,|\psi|]=\frac{1}{|G|}\sum_{g \in G}|\chi(g)|\overline{|\psi(g)|}=\frac{1}{|G|}\sum_{g \in G}|\chi(g)||\psi(g)|$ is a non-negative (even positive, but we do not need that now) real number. Now applying the inequality of Cauchy-Schwartz (see for example here) we have
$$\Big(\sum_{g \in G}|\chi(g)||\psi(g)|\Big)^2 \leq \sum_{g \in G}|\chi(g)|^2 \cdot \sum_{g \in G}|\psi(g)|^2$$
and since $\chi$ and $\psi$ are irreducible, $[\chi,\chi]=1=[\psi,\psi]$, so the above inequality amounts to $\Big(\sum_{g \in G}|\chi(g)||\psi(g)|\Big)^2 \leq |G|^2$ from which $[|\chi|,|\psi|] \leq 1$ follows.$\square$


Proposition Let $\chi, \psi, \zeta \in Irr(G)$, then $[\chi\psi,\zeta] \leq min(\chi(1),\psi(1),\zeta(1)).$

Proof Note that the product of two characters is again a character, hence the inner product $[\chi\psi,\zeta]$ is a non-negative integer, and it suffices to show that $[\chi\psi,\zeta] \leq \zeta(1)$, since complex conjugates of irreducibles are again irreducible and $[\chi\psi,\zeta]=[\chi,\overline{\psi}\zeta]=\overline{[\chi,\overline{\psi}\zeta]}=[\overline{\psi}\zeta, \chi]=[\chi\overline{\zeta},\overline{\psi}]$. Now, using the triangle inequality, the fact that $|\zeta(g)| \leq \zeta(1)$ (see I.M. Isaacs, Character Theory of Finite Groups, Lemma (2.15)(c)), and the Lemma above, we get
$$[\chi\psi,\zeta]=|[\chi\psi,\zeta]|=\frac{1}{|G|}|\sum_{g \in G}\chi(g)\psi(g) \overline{\zeta(g)}| \leq \frac{1}{|G|}\zeta(1)\sum_{g \in G}|\chi(g)||\psi(g)|=\zeta(1)[|\chi|,|\psi|] \leq \zeta(1)$$ and we are done.
