Convergence in the cut norm implies convergence in L2

[[concept]]

Theorem

Let $W:[0,1{]}^2\to [-1,1]$. Then $||W|{|}_{2,2}\stackrel{(1)}{\le }\sqrt{4\cdot ||W|{|}_{\infty \to 1}}\stackrel{(2)}{\le }\sqrt{16\cdot ||W|{|}_{\square }}$ ie, cut norm convergence implies convergence in $L_2$

Proof of (2)

We first show inequality (2). Using the definition of the cut norm, we have

\lvert \lvert W \rvert \rvert_{\square} &= \sup_{S,T \subseteq [0,1]} \left\lvert \int \int _{S \times T} W(u,v) \, du \, dv \right\rvert \\ &= \sup_{f,g: [0,1]\to[0,1]} \left\lvert \int_{0}^1 \int_{0}^1 W(u,v) f(u) g(v) \, du \, dv \right\rvert \\ (*) &= \sup_{f,g: [0,1] \to [0,1]}\lvert \langle T_{W}f, g \rangle \rvert \\ \end{aligned}$$ Since we are taking the supremum, it is equivalent to taking the supremum over all $L_{\infty}$ functions $f$ and $g$. To get $(*)$, we notice that $\int _{0}^1 W(u,v) f(u) \, du$ is an [[Concept Wiki/integral linear operator]] with kernel $W$. We denote this with $T_{W}$ so that $$T_{W}f\,\,(v) =\int _{0}^1 W(u,v) f(u) \, du$$ - or equivalently that $T_{W}f = \int _{0}^1 W(u,\cdot) f(u) \, du$ We can then see that the definition becomes the absolute value of the inner product of two $L_{\infty}$ functions. Note that $\lvert \lvert W \rvert \rvert_{\infty \to 1}$ is an induced [[Concept Wiki/operator norm]] of the [[Concept Wiki/graphon]] mapping functions from $L_{\infty}$ to $L_{1}$. $$\begin{aligned} \lvert \lvert W \rvert \rvert_{\infty \to 1} &= \sup_{-1 \leq g \leq 1} \lvert \lvert T_{W}g \rvert \rvert_{1} \\ &= \sup_{-1 \leq g \leq 1} \int \lvert T_{W}g(x) \rvert \, dx \\ &= \sup_{-1 ≤f,g≤1} \langle T_{W}g, f \rangle \\ \end{aligned}$$ - we omit the denominator in our usual definition of the [[Concept Wiki/operator norm]] since $g$ is bounded by $-1$ and $1$ and $g \in L_{\infty}$ - We can rewrite this using $f$ to match in the definition of the $L_{1}$ norm since we are taking the supremum - For more about $L_{p}$ spaces see [wikipedia](https://en.wikipedia.org/wiki/Lp_space#Lp_spaces_and_Lebesgue_integrals) (hand wavy/cursory for this class on functional analysis details) We can then rewrite this as $$\begin{aligned}\lvert \lvert W \rvert \rvert_{\infty \to 1} &= \sup_{-1 \leq f, f' \leq 1, -1 \leq g, g' \leq 1} \langle T_{W}(g-g'), f-f' \rangle \\ (**) &\leq \sup_{-1 \leq g, f \leq 1} \langle T_{W}g, f \rangle - \sup_{-1 \leq g,f' \leq 1} \langle T_{W}g, f' \rangle + \sup_{-1 \leq g',f\leq 1} \langle T_{W}g', f \rangle - \sup_{-1 ≤g', f' ≤1} \langle T_{W}g', f' \rangle \\ ({\dagger}) &\leq 4 \lvert \lvert W \rvert \rvert_{\square} \end{aligned}$$ Where we get $(**)$ by the [[Concept Wiki/triangle inequality]] and $({\dagger})$ from the definition of $\lvert \lvert \cdot \rvert \rvert_{\square}$. - Note that this is a loose bound! And this gives us the desired result for (2). ^proof-2

Proof of (1)

Proving (1) is more involved and uses more functional analysis.

Use the Riesz-Thorin interpolation theorem for complex $L_p$ spaces: $||W|{|}_{p\to q}\le ||W|{|}_{p_1\to q_0}^{1-\theta }||W|{|}_{p_1\to q_1}^{\theta }$ Where

• $\theta =\min \left(1-\frac{1}{p},\frac{1}{q}\right)$ and
• $p_0,q_0\in [1,\infty )$
• with $\frac{1}{p}=\frac{1-\theta }{p_0}$ and
• $1-\frac{1}{q}=(1-\theta )\left(1-\frac{1}{q_0}\right)$ and
• $p_1=\infty ,q_1=1$.

Define operator norm$||W|{|}_{\square ,\mathbb{C}}={\sup }_{\begin{array}{rl}f,g:[0,1]& \to \mathbb{C}\\ ||f|{|}_{\infty },||g|{|}_{\infty }& \le 1\end{array}}\left|{\int }_0^1{\int }_0^{1}W(u,v)f(u)g(v)dudv\right|$ So for complex functions, we can see that

\text{for complex functions } \lvert \lvert W \rvert \rvert_{\infty \to 1} &= \lvert \lvert W \rvert \rvert_{\square, \mathbb{C}} \\ &≤ 2\lvert \lvert W \rvert \rvert_{\infty \to 1} \text{ for real functions} \end{aligned}$$ We then have $$\lvert \lvert W \rvert \rvert _{p_{0} \to q_{0}} \leq \lvert \lvert W \rvert \rvert _{1 \to \infty} \leq \lvert \lvert W \rvert \rvert_{\infty} \leq 1$$ Since $W \leq 1$, and this gives the desired result.

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