eigenvalues of the induced graphon converge pointwise to the eigenvalues of the limit

[[concept]]

Theorem

Let be a convergent graph sequence with limit graphon . Then ie, the eigenvalues of the induced graphon converge to the eigenvalues of the limit.

Where

Proof

Thus

T_{W_{n}} \varphi(v) &= \int _{0}^1 W_{n}(u,v) \varphi(u)\, du &= \lambda \varphi(v) \\ &= \int _{0}^1 \sum_{i,j=1}^n [A_{n}]_{ij} \mathbb{1}(u \in I_{i}) \mathbb{1}(v \in I_{j}) \varphi(u)\, du &= \lambda \varphi(v) \\ &= \mathbb{1}(v \in I_{j}) \int _{0}^1 \sum_{i,j=1}^n [A_{n}]_{ij} \mathbb{1}(u \in I_{i}) \varphi(u) \, du &= \lambda \varphi(v) \\ \implies &= \sum_{i,j=1}^n [A_{n}]_{ij} \int _{0}^1 \mathbb{1}(u \in I_{i}) \varphi(u) \, du &=\lambda \varphi(v \in I_{j}) \end{aligned} $$ LHS is a constant for each $v \in I_{j}$. This means that $\varphi$ is a step function $\varphi(v \in I_{j}) = x_{j}$. Thus, we can rewrite the last line as $$\begin{aligned} \lambda x_{j} &= \sum_{i=1}^n [A_{n}]_{ij} x_{i} \frac{1}{n} \end{aligned}$$ since for each $i$ we have $\varphi(u) = \begin{cases} x_{i},\,\,x_{i} \in I_{i}\\0, \,\,\,\text{otherwise}\end{cases}$ . The above is just a matrix multiplication, so we get $$\begin{align} \implies \lambda x &= \frac{1}{n} A_{n}x \\ \implies \lambda_{j} (T_{W_{n}}) &= \frac{\lambda_{j}(A_{n})}{n} \;\;\;\forall j \end{align}$$ For convergence, this means that $$\begin{align} (\text{1}) & &\, t(c_{k}, G_{n}) &= \sum_{i=1}^n \lambda_{i}^k \\ (2) & &\, t(c_{k}, G_{n}) &\to \tilde{t}(c_{k}, W) \end{align}$$ where - $c_{k}$ is the $k$-cycle graph - $t$ is the graph [[Concept Wiki/homomorphism density]] - $\tilde{t}$ is the [[Concept Wiki/graphon homomorphism density]] (the full proof involves writing out the expressions for (1) and showing (2) that the elements in the sum converge pointwise. We do not show this for this class.)

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