This is related to my [question](https://mathoverflow.net/q/314844), but it concerns a specific point of the proof of Schur's Theorem.


Let $G$ be a finite group and $\chi$ an irreducible character of $G$. Is it true that
$$\forall g\in G,\qquad\sum\_{h\in G}\overline{\chi(h)}\chi(gh)=\frac{|G|}{\chi(e)}\,\chi(g)\quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $\chi$ is a linear character, or if $g=e$. I checked it for $G=\frak S\_3$ and for one nonlinear character of $\frak S\_4$.



A character identity
Yes. Let 
$$ F(g) = \sum\_{h \in G} \overline{\chi(h)}\chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^{-1} g s) = \sum\_{h \in G} \overline{\chi(h)}\chi(s^{-1}gsh)=\sum\_{h \in G} \overline{\chi(h)}\chi(g(shs^{-1})) = \sum\_{h' \in G} \overline{\chi(s^{-1}h's )}\chi(gh') = \sum\_{h' \in G} \overline{\chi(h' )}\chi(gh') = F(g).$$


Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = \sum\_{i} a\_{\chi\_i} \chi\_i(g).$$
Let $\rho\_{\chi\_i}$ be the representation with character $\chi\_i$. Then
$$a\_{\chi\_i} = \frac{1}{|G|} \sum\_{g \in G} \overline{\chi\_i(g)}F(g) = \frac{1}{|G|} \sum\_{g,h} \overline{\chi\_i(g)} \overline{\chi(h)}\chi(gh)= \frac{1}{|G|} \sum\_{h} \overline{\chi(h)} (\mathrm{Tr}( [\sum\_{g} \overline{\chi\_i(g)}\rho\_{\chi}(g)]\rho\_{\chi}(h) ). $$
Letting
$$A\_i = \sum\_{g} \overline{\chi\_i(g)}\rho\_{\chi}(g),$$
we have $$(\*)a\_{\chi\_i} = \frac{1}{|G|} \sum\_{h} \overline{\chi(h)} (\mathrm{Tr}( A\_i\rho\_{\chi}(h) ).$$ 
I claim $A\_i$ commutes with $\rho\_{\chi}(h)$ for all $h$-s:
$$A\_i \rho\_{\chi}(h) =\sum\_{g} \overline{\chi\_i(g)}\rho\_{\chi}(gh) =\sum\_{g'} \overline{\chi\_i(g'h^{-1})}\rho\_{\chi}(g')=\sum\_{g'} \overline{\chi\_i(h^{-1} g')}\rho\_{\chi}(g')=\sum\_{g} \overline{\chi\_i(g)}\rho\_{\chi}(hg)= \rho\_{\chi}(h)A\_i.$$
Thus, by Schur's lemma, $A\_i$ is a multiple of the identity:
$$A\_i = c\_i I$$
for some constant $c\_i$, which can be extracted by taking traces:
$$c\_i \cdot \dim(\rho\_i) = \mathrm{Tr}(A\_i) = \sum\_{g \in G}\overline{\chi\_i}(g) \chi(g) = 1\_{\chi\_i = \chi} |G|.$$ 
Thus, when $\chi\_i \neq \chi$ we have $A\_i=0$ and so $a\_{\chi\_i}=0$ in that case by $(\*)$, and $F(g)$ must be proportional to $\chi(g)$. To find the constant of proportionality, note that if $\chi\_i=\chi$, we have from $(\*)$ that
$$a\_{\chi\_i}=\frac{1}{\dim \rho\_{\chi\_i}}\sum\_{h} \overline{\chi\_i}(h) \chi(h) = \frac{|G|}{\chi(e)},$$
as needed.


