Convergent sum of operator norms of iterates of kernel operator corresponding to stepping of graphon $W$

Question

Suppose that we are given a possibly non-symmetric graphon $W:[0,1]^2\to\mathbb [0,K]$, where $K>0$ is some fixed, known constant. Given such a graphon, we consider the kernel operator $T_W:L^1[0,1]\to L^1[0,1]$ defined by $$(T_Wf)(\cdot)=\int_0^1W(\cdot,y)f(y)\ \mathrm dy;$$ see $\S$7.5 of Lovasz - Large networks and graph limits.

Now assume that $W$ is such that $\rho(T_W)<1$, i.e. the spectral radius of the operator $T_W$ is smaller than one. By a straightforward application of Gelfand's formula, we can prove that $$\sum_{n\geq0}\|T_W^n\|<\infty,$$ using the exponential decay of $\|T_W^n\|$.

Now suppose that we are given a sequence of partitions $(\mathcal P^d)_{d\in\mathbb N}$ of $[0,1]$ into consecutive intervals $I_1,\ldots,I_d$, in such a way that $\mathrm{mesh}(\mathcal P^d)\to0$ as $d\to\infty$. I'm willing to assume that $\mathcal P^{d+1}$ is a refinement of $\mathcal P^d$, but I don't think this is of much help. Now consider the stepping of $W$, denoted by $W_{\mathcal P^d}$, obtained by averaging over the elements of the partition $\mathcal P^d$. In other words, we have $$W_{\mathcal P^d}(x,y)=\sum_{i=1}^d\sum_{j=1}^d\mathbf1\{x\in I_i,y\in I_j\}\frac1{\lambda(I_i)\lambda (I_j)}\int_{I_i\times I_j}W(x,y)\ \mathrm dx\ \mathrm dy,$$ where $\lambda$ denotes the Lebesgue measure. See $\S$9.2 of Lovasz - Large networks and graph limits. Of course, we can also consider the kernel operators $T_{W_{\mathcal P^d}}:L^1[0,1]\to L^1[0,1]$ associated to those graphons. Note that the stepping graphons correspond to finite dimensional projections of $W$, i.e. to graphs.

I'm able to prove that given $\rho(T_W)<1$ and some $\epsilon>0$ such that $\rho(T_W)+\epsilon<1$, we can find some $d^*$ such that for all $d\geq d^*$, $\rho(T_{W_{\mathcal P^d}})<\rho(T_W)+\epsilon$. By the same arguments as before, this implies that $$\sum_{n\geq0}\|T_{W_{\mathcal P^d}}^n\|<\infty,$$ for any $d\geq d^*$. However, in the application I have in mind, it would be helpful for me to have an uniform bound on those summations, i.e. I want to find some $C>0$ such that for all $d\geq d^*$ (or if needed, for all $d\geq d^{**}$ for some large $d^{**}$) it holds that $$\sum_{n\geq0}\|T_{W_{\mathcal P^d}}^n\|\leq C.$$ I expect such a bounding to hold, since to prove such a thing, we need (i) eventually exponential decay of the norms, which can be assured by $\rho(T_{W_{\mathcal P^d}})<\rho(T_W)+\epsilon<1$; and (ii) that this exponential decay 'kicks in' sufficiently fast. For this second condition, I believe that the steppings $T_{W_{\mathcal P^d}}$ 'mix faster' than the operators $T_W$ since they are piecewise constant, and do not use information within the intervals $I_i$. In particular, if we find $N$ such that for $n\geq N$ it holds that $\|T_W^n\|<((1+\alpha)\rho(T_W))^n$, then I would expect that $\|T_{W_{\mathcal P^d}}^n\|\leq((1+\alpha)(\rho(T_W)+\epsilon))^n$ as well. Then choose $\alpha$ sufficiently small, fix this $N$, bound the first $N$ terms, and then bound the final part of the summation using Gelfand's formula. However, I find it difficult to make this argument formal.

I was hoping the Uniform Boundedness Principle could be of help, but I wasn't able to prove my conjecture in this way either.

It seems that when $W$ is symmetric (an assumption I wish to avoid), then since $T_W$, as well as the stepping operators, are self-adjoint, we could use spectral theory and e.g. Davis-Kahan theory to prove such a thing, by showing that the eigenvector corresponding to the largest eigenvalue of $T_{W_{\mathcal P^d}}$ converges to that of $T_W$; at least if we work on $L^2[0,1]$. However, I do not assume $W$ to be symmetric, and I'm looking for a different approach.

Any help or reference is much appreciated. Feel free to impose more assumptions if you believe them to be necessary, but I do not wish to assume that $W$ is symmetric.

Answer 1

The sums are indeed uniformly bounded. For legibility I'll write $T_d$ instead of $T_{W_{\mathcal P^d}}.$ I'll use the notation $\|X\|_{p\to q}$ for $\sup\{\|X(f)\|_q:\|f\|_p=1\}$ with $p,q\in\{1,\infty\}.$ Here are the main ideas/observations:

1. by submultiplicativity + bounding by a geometric series, it suffices to show $\|T_d^n\|\to\|T^n\|$ for each fixed $n$
2. $\|T_d^n\|\leq \|T_d^{n-1}T_W\|$
3. $\|T_d-T_W\|_{\infty\to 1}\to 0$
4. $\|T_d\|_{p\to q},\|T_W\|_{p\to q}\leq K$ for all $p,q\in\{1,\infty\}$

In more detail:

1. there exists $\epsilon>0$ and $N$ such that $\|T^N\|\leq 1-\epsilon.$ Assume we can show that $\|T_d^n\|\leq \|T^n\|+\epsilon/2$ for $n\leq N$ and sufficiently large $d.$ Then \begin{align*} &\sum_{n\geq 0}\|T_d^n\|\\ &= \sum_{m\geq 0}\sum_{n<N}\|T_d^{Nm+n}\|\\ &\leq \sum_{m\geq 0}\|T_d^{N}\|^m\sum_{n<N}\|T_d^n\|\\ &\leq \sum_{m\geq 0}(1-\epsilon/2)^m\sum_{n<N}(\|T^n\|+\epsilon/2) \end{align*} which is finite and independent of $d.$

2. we can write $T_d=S_dT_WS_d$ where $S_d$ is the "stepping operator" $L^1\to L^1$: the conditional expectation operator onto the $\sigma$-algebra generated by $\mathcal P_d.$ Since $S_d$ is a contraction in $L^1,$ the $L^1\to L^1$ norm of $T_d^{n-1}S_dT_WS_d$ is at most the $L^1\to L^1$ norm of $T_d^{n-1}S_dT_W=T_d^{n-1}T_W.$

3. This is the obvious estimate $\|T_X\|_{\infty\to 1}\leq\|X\|_1$ with $X=W_{\mathcal P^d}-W,$ and the fact that $W_{\mathcal P^d}$ tends to $W$ in $L^1([0,1]^2)$ (by the Lebesgue density theorem for example).

4. easy

Now to put it all together. By 1 and 2 it suffices to show that for all $n$ we have $\|T_d^{n-1}T_W-T_W^n\|_{1,1}\to 0$ as $d\to\infty.$ This follows from the telescoping sum $T_d^{n-1}T_W-T_W^n=\sum_{m=1}^{n-1}T_d^{n-m-1}(T_d-T_W)T_W^m$ and \begin{align*} &\|T_d^{n-m-1}(T_d-T_W)T_W^m\|_{1\to 1}\\ &\leq\|T_d\|_{1\to 1}^{n-m-1} \|T_d-T_W\|_{\infty\to 1} \|T_W\|_{\infty\to\infty}^{m-1} \|T_W\|_{1\to \infty}\\ &\leq K^{n-1} \|T_d-T_W\|_{\infty\to 1}&\qquad\text{(by 4)}\\ &\to 0&\qquad\text{(by 3)} \end{align*}