Convergence in the cut norm implies convergence in L2

Let $W:[0,1{]}^2\to [-1,1]$. Then

ie, cut norm convergence implies convergence in $L_2$

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

Since we are taking the supremum, it is equivalent to taking the supremum over all $L_{\mathrm{\infty }}$ functions $f$ and $g$.

To get $(\ast )$, we notice that ${\int }_0^{1}W(u,v)f(u)du$ is an integral linear operator with kernel $W$. We denote this with $T_W$ so that

• 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 $L_{\mathrm{\infty }}$ functions.

Note that $||W|{|}_{\mathrm{\infty }\to 1}$ is an induced operator norm of the graphon mapping functions from $L_{\mathrm{\infty }}$ to $L_1$.

• we omit the denominator in our usual definition of the operator norm since $g$ is bounded by $-1$ and $1$ and $g\in L_{\mathrm{\infty }}$
• 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 (hand wavy/cursory for this class on functional analysis details)

We can then rewrite this as

Where we get $(\ast \ast )$ by the triangle inequality and $(†)$ from the definition of $||\cdot |{|}_{◻}$.

• Note that this is a loose bound!

And this gives us the desired result for (2).
^proof-2

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

Use the Riesz-Thorin interpolation theorem for complex $L_p$ spaces:

Where

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

Define operator norm$$\lvert \lvert W \rvert \rvert_{\square, \mathbb{C}} = \sup_{\begin{aligned}
f,g:[0,1] &\to \mathbb{C} \\lvert \lvert f \rvert \rvert_{\infty}, \lvert \lvert g \rvert \rvert_{\infty} &≤1 \end{aligned}} \left\lvert \int {0}^1 \int^1 W(u,v) f(u) g(v) , du , dv \right\rvert $$
So for complex functions, we can see that

We then have

Since $W\le 1$, and this gives the desired result.