Fourier Inversion Theorem and Plancherel's Identity in the context of Representations This page (https://en.wikipedia.org/wiki/Fourier_transform_on_finite_groups) supplies some properties of the Fourier transform that I am having a difficult time trying to prove. 
In particular, given the Fourier transform of $\phi$ at a representation $(\rho,V)$ defined by $\widehat{\phi}(\rho) = \sum_{g\in G}\phi(g)\rho(g)$, I first want to prove the Fourier inversion theorem given by $$\phi(g) = \frac{1}{|G|}\sum_{\rho\in\text{Irrep(G)}}\deg(\rho)\text{tr}(\rho(g^{-1})\widehat{\phi}(\rho))$$ and then make use of the Fourier inversion theorem to prove the Plancherel identity $$\sum_{g\in G}\phi_1(g^{-1})\phi_2(g) = \frac{1}{|G|}\sum_{\rho\in\text{Irrep(G)}}\deg(\rho)\text{tr}(\widehat{\phi_1}(\rho)\widehat{\phi_2}(\rho)).$$
I have been trying at these for a few days now but haven't been able to prove these due to lacking knowledge in character theory, so I would be immensely appreciative if anyone could supply proofs of these.
Edit: I would seriously be so appreciative if someone could provide one or both of these proofs!
 A: I'll explain how I think about characters, then answer your question.
If $\rho,\theta$ are irreps, then 
$$\dim \text{Hom}_{G\text{ reps}}(\rho,\theta) \ = \ \begin{cases} 0& \text{ if }\rho\not=\theta\\
1&\text{ if }\rho=\theta
\end{cases}$$
so $\dim \text{Hom}_G(-,-)$ is a sort of inner product on representations. To make this precise, remember that for any representations $\rho,\theta$
$$\text{Hom}_{G\text{ reps}}(\rho,\theta)\ = \ \text{Hom}_{\text{vect}.}(\rho,\theta)^G$$
If $V$ if a vector space with a $G$ action, Then 
$$\pi \ :\ V\ \longrightarrow \ V \ \ \ \text{ sending } \ \ \ v\ \longrightarrow \ \frac{1}{|G|}\sum_g g\cdot v$$
is a projection onto the $G$ invariants. Since $\pi^2=\pi$, its trace is the dimension of its image. Applying this to $V=\text{Hom}_{\text{vect}.}(\rho,\theta)$, we get
$$ \dim V^G\ = \ \text{tr}\pi \ = \ \frac{1}{|G|}\sum_g \text{tr}(\rho(g^{-1})\theta(g))$$
This is where the formula for $\dim \text{Hom}_{G\text{ reps}}(\rho,\theta)$ (for arbitrary representations $\rho,\theta$) comes from. It gives an inner product $(-,-)$ on representations.

Now to the proofs of Fourier inversion (1) and the Plancherel identity (2).


*

*The right side is the function $$g\ \longrightarrow \ \frac{1}{|G|}\sum_\rho\sum_h\dim\rho\ \text{tr} \rho(g^{-1}h) \phi(h)\ = \ \frac{1}{|G|}\sum_\rho\sum_h \overline{\text{tr}\rho(1)} \text{tr}\rho(h)\phi(gh)\ = \ \phi(g)$$
where the last equality is by column orthogonality, since only $1\in G$ is conjugate to $1$.

*Let $\phi_1,\phi_2$ be the indicator functions at $g_1,g_2$. The general case follows by linearity. The right side is $$\frac{1}{|G|}\sum_\rho \dim\rho \ \text{tr}(\rho(h_1)\rho(h_2))\ = \ \frac{1}{|G|}\sum_\rho\overline{\text{tr}\rho(1)}\text{tr}(\rho(h_1h_2)) $$ This is $0$ if $h_1\ne h_2^{-1}$ by column orthogonality, similarly $1$ if $h_1=h_2^{-1}$. So this is the same as the left hand side.
