Spectral Graph Convolutions: What are the spectral filters functions I am trying to understand the mathematical meaning of one of the steps that appear in the Convolution Theorem (Step 4 here).
To give some context, this is related to applying the convolution theorem with Graphs. For that it uses the Eigen Decomposition of the normalized Laplacian Matrix from which the Fourier Functions / Fourier Transform appear as the EigenVectors of this Matrix.
So elaborating all the concepts and giving context:
Graph Laplacian $\Delta$
$$ \begin{aligned} & \mathcal{G}=(V, E, A)  \rightarrow \underset{n \times n}{\Delta}=\mathrm{I}-D^{-1 / 2} A D^{-1 / 2}  \\ 
& \text { where } \\ 
& \quad \quad \underset{n \times n} D=\operatorname{diag}  (\sum_{j \neq i} A_{i j}) \\
& \quad \quad A \equiv \text adjacency \quad Matrix
\end{aligned} $$
Note: This is called the normalized Laplacian Matrix


Fourier Functions
The following is the eigen-decomposition of graph Laplacian,
$$ \underset{n \times n}{\Delta}=\Phi^{T} \Lambda \Phi $$
$\Phi$ contains the Lap eigenvectors $\phi_1$ to $\phi_n$, each of size $n \times 1$, and those are also called Fourier functions that form an orthonormal basis ($\Delta$ is positive semi-definite matrix) so  $\begin{array}{ll} \underset{n \times 1}{\Delta \phi_{k}}=\lambda_{k} \phi_{k}, & k=1, \ldots, n \ & \end{array}$.
Then my understanding is that $\phi_i$ is a vector and not a function (this is relevant for the question below)


Fourier Transform and Inverse Fourier Transform
The Fourier transform is basically projecting a function h on the Fourier functions, and the result are the coefficients of the Fourier series
Fourier Transform/ coefficients of Fourier Series
$$ \underset{n \times 1}{\mathcal{F}(h)} =\Phi^{T} h \ =\hat{h} $$
Inverse Fourier Transform
$$ \underset{n \times 1}{\mathcal{F}^{-1}(\hat{h})} =\Phi \hat{h} = \Phi \Phi^{T} h=h $$
Convolution Theorem
Fourier transform of the convolution of two functions is the pointwise product of their Fourier transforms. Below is stated the theorem where $\Phi^{T}w = \hat{w}$
$$ \begin{aligned} \underset{n \times 1} {w * h} &=\mathcal{F}^{-1} (\mathcal{F}(w) \odot \mathcal{F}(h))) \\ &=\underset{n \times n}{\Phi}\left( \underset{n \times 1}{\hat{w}}\odot \underset{n \times 1}{\Phi^{T} h}\right) \\ &=\Phi\left(\underset{n \times n}{\hat{w}(\Lambda)} \underset{n \times 1}{\Phi^{T} h}\right) \\ &=\Phi \hat{w}(\Lambda) \Phi^{T} h \\ &=\hat{w}(\Phi \Lambda \Phi^{T}) h \\ &=\underset{n \times n}{\hat{w}(\Delta)} \underset{n \times1}h \ & \end{aligned} $$
With all this context, now my question is what is the $\hat{w}$ vector that is treated like a function?
In the third $=$ of the convolution theorem you can see that the pointwise multiplication has been replaced by a matrix multiplication converting $\hat{w}$ to $\hat{w}(\Lambda)$. This is understandable to me interpreting $\hat{w}$ as a function instead of a vector then each of the components of $\hat{w}$ correspond to evaluate the function on one of the eigenvalues $\lambda_i$, and these are then only points you have of this function:
$$    
\hat{w} = \begin{bmatrix}
           \hat{w}(\lambda_1) \\
           \hat{w}(\lambda_2) \\
           \vdots \\
           \hat{w}(\lambda_n)
        \end{bmatrix} 
$$
So it is understandable that:
$$
        \text Given \quad \Lambda =  \begin{bmatrix}
           \lambda_1 & 0 & \cdots & 0 \\           
           0 & \lambda_2 & \cdots & 0 \\
           \vdots & \vdots & \vdots & \vdots \\
           0 & 0 & \cdots & \lambda_n \\
          \end{bmatrix} \implies  \hat{w}(\Lambda) =
          \begin{bmatrix}
           \hat{w}(\lambda_1) & 0 & \cdots & 0 \\           
           0 & \hat{w}(\lambda_2) & \cdots & 0 \\
           \vdots & \vdots & \vdots & \vdots \\
           0 & 0 & \cdots & \hat{w}(\lambda_n) \\
          \end{bmatrix}
$$
This is clear and makes sense because the vector $\hat{w}$ is by construction defined in the eigenvalues. But what about the 5th and 6th $=$? How do you interpret $\hat{w}(\Phi \Lambda \Phi^{T})$ and $\hat{w}(\Delta)$?
There is also a great video on Youtube in https://www.youtube.com/watch?v=Iiv9R6BjxHM&t=2985s where you can go over the explanation of this (but it does not explain what I am asking, Xavier just refers to it as the spectral filter) on the minute 40:13 and 46:52

I think this is a case of sloppy / inconsistent / informal notation. $\hat w(\Lambda)$ is just the vector $\hat w$ rearranged into a diagonal matrix. Then you could say, "actually let's call the entire thing $\Phi \text{diag}(\hat w) \Phi^T$ just $\hat w(\Delta)$ for short. And then, because we want to make $\hat w(\Delta)$ some learnable function with a fixed number of parameters and a limited computational budget, let's actually define it as, for example, $\hat w(\Delta) = \theta_1 \Delta + \theta_2 \Delta^2$ (this is the "LapGCN" example from the next page of the linked article).
