On a representation theory course, we were given the following Proposition:
Let $G$ be a finite group, $H$ a subgroup and $χ_{H,1},\dots,χ_{H,r}$ the irreducible characters of $H$. Let also $χ$ be a character of $G$.
If $Res_H^G(χ)=\sum\limits_{i=1}^{r}a_{i}χ_{H,i}$, for some $a_i\in\mathbb{N}$, then: $\sum\limits_{i=1}^{r}a_{i}^{2}\leq[G:H]$
Now, by the way this proposition is phrased, it means that the above bound holds for $\textbf{any}$ character of $G$, both irreducible and non-irreducible, but I believe that this is wrong.
I think the above statement holds only when $χ$ is an $\textbf{irreducible}$ character of $G$.
If we want to make the above statement hold for any character of $G$, then I believe it should be stated as:
Let $G$ be a finite group, $H$ a subgroup and $χ_{G,1},\dots,χ_{G,s},χ_{H,1},\dots,χ_{H,r}$ the irreducible characters of $G$ and $H$ respectively. Let also $χ$ be a character of $G$.
If $χ=\sum\limits_{i=1}^{s}b_{i}χ_{G,i}$ and $Res_H^G(χ)=\sum\limits_{i=1}^{r}a_{i}χ_{H,i}$, for some $a_i,b_i\in\mathbb{N}$, then: $\sum\limits_{i=1}^{r}a_{i}^{2}\leq[G:H]\sum\limits_{i=1}^{s}b_{i}^{2}$
and I prove it as follows:
We have:
$$\langle Res_H^G(χ),Res_H^G(χ)\rangle_{H} = \langle\sum\limits_{i=1}^{r}a_{i}χ_{H,i},\sum\limits_{j=1}^{r}a_{j}χ_{H,j}\rangle_{H}$$
For the right hand side of the equality: $$\langle\sum\limits_{i=1}^{r}a_{i}χ_{H,i},\sum\limits_{j=1}^{r}a_{j}χ_{H,j}\rangle_{H} = \sum\limits_{i=1}^{r}\sum\limits_{j=1}^{r}a_{i}a_{j}\langle χ_{H,i},χ_{H,j}\rangle_{H}=\sum\limits_{i=1}^{r}a_{i}^{2}$$ where we used the orthogonality of the irreducible characters $χ_{H,i}$.
For the left hand side of the equality: $$\langle Res_H^G(χ),Res_H^G(χ)\rangle_{H}=\frac{1}{|H|}\sum\limits_{h\in H}|Res_H^G(χ)(h)|^2\leq \frac{1}{|H|}\sum\limits_{g\in G}|Res_H^G(χ)(g)|^2 = \frac{|G|}{|H|}\frac{1}{|G|}\sum\limits_{g\in G}|χ(g)|^2 = [G:H]\langle χ,χ \rangle_{G}$$ where the inequality is justified since we are adding more non-negative terms, and we identified $Res_{H}^{G}(χ)$ with $χ$ since we are "running through" the elements of the entire group $G$ rather than just $H$.
Now: $$\langle χ,χ \rangle_{G} =\langle\sum\limits_{i=1}^{s}b_{i}χ_{G,i},\sum\limits_{j=1}^{s}b_{j}χ_{G,j}\rangle_{G}= \sum\limits_{i=1}^{s}\sum\limits_{j=1}^{s}b_{i}b_{j}\langle χ_{G,i},χ_{G,j}\rangle_{G} = \sum\limits_{i=1}^{s}b_{i}^{2}$$ where again we used the orthogonality of the $χ_{G,i}$
Putting all these together we get the desired result, namely: $$\sum\limits_{i=1}^{r}a_{i}^{2}\leq[G:H]\sum\limits_{i=1}^{s}b_{i}^{2}$$
Is the alternative statement, along with the proof I provided correct, or am I missing something regarding the Proposition we were given in the lecture?