Having read Lemma 7 (duality lemma) for quadratic forms, I wonder if a similar result holds for (the more general) bilinear sums. That is, does the following claim hold? First for a sequence of (say finite or square-summable) complex numbers $\alpha_n$, let $\left\| \alpha\right\| := (\sum_{n}|\alpha_n|^2)^{1/2}$.
Claim: Let $\phi(m,n)$ be a function on $\mathbb{Z}^2$ and suppose there is a constant $\Delta$ such that for all complex numbers $\gamma_n, \delta_n$, we have $$ \left| \sum_{m} \sum_{n} \sum_{n'} \gamma_n \overline{\delta_{n'}} \phi(m,n) \overline{\phi(m,n')}\right| \leq \Delta \left\| \gamma\right\|\left\| \delta\right\|. $$ Then for all complex numbers $\alpha_m, \beta_m$, we have $$ \left| \sum_n \sum_m \sum_{m'} \alpha_m \overline{\beta_{m'}} \phi(m,n) \overline{\phi(m',n)}\right| \leq \Delta \left\| \alpha\right\| \left\| \beta \right\|. $$ Here $n,n'$ run over some set $\mathcal{N}$ and $m,m'$ run over some set $\mathcal{M}$.
A try: I tried the following but became stuck towards the end. Perhaps there is an idea at the end that could finish the proof, but I'm not sure.. Here it is:
Define the complex numbers $\gamma_n := \sum_{m}\overline{\beta_m \phi(m,n)}$. By Cauchy-Schwarz, \begin{align*} |\Phi|^2 &:= \left| \sum_n \sum_m \sum_{m'} \alpha_m \overline{\beta_{m'}} \phi(m,n) \overline{\phi(m',n)}\right|^2 \leq \sum_m |\alpha_m|^2 \sum_m\left| \sum_n\phi(m,n)\sum_{m'} \overline{\beta_{m'}\phi(m',n)}\right|^2\\ &= \sum_m |\alpha_m|^2 \sum_m\left| \sum_n \gamma_n \phi(m,n)\right|^2 = \left\| \alpha \right\|^2\sum_{m}\sum_{n} \sum_{n'}\gamma_n \overline{\gamma_{n'}} \phi(m,n) \overline{\phi(m,n')}\\ &\leq \left\| \alpha \right\|^2 \Delta \left\| \gamma \right\|^2. \end{align*} The last inequality holds by the hypothesis. Now it only remains to show that $\left\| \gamma \right\|^2 \leq \Delta \left\| \beta \right\|^2$. We have, by Cauchy-Schwarz, \begin{align*} \left\| \gamma \right\|^2 &= \sum_n | \gamma_n|^2 = \sum_n \left| \sum_m\overline{\beta_m \phi(m,n)} \right|^2 \leq \sum_n \sum_{m}|\beta_m|^2 \sum_{m} |\phi(m,n)|^2\\ &= \left\| \beta \right\|^2 \sum_m \sum_n |\phi(m,n)|^2. \end{align*} We need to show that $\sum_m \sum_n |\phi(m,n)|^2 \leq \Delta$. Here is where I get stuck. Most certainly we should use the hypothesis again here. That is, we should construct complex numbers $y_n, z_n$ with $\left\| y\right\| = \left\| z\right\| = 1$ and such that $$ \sum_m \sum_n |\phi(m,n)|^2 \leq \left|\sum_m \sum_n \sum_{n'}y_n \overline{z_{n'}} \phi(m,n) \overline{\phi(m,n')} \right|. $$ The result would then follows from the hypothesis.
One could try here something like \begin{align*} \sum_m \sum_n |\phi(m,n)|^2 &= \int_{0}^1\sum_m \sum_n \sum_{n'}e(nx) \overline{e(n'x)}\phi(m,n) \overline{\phi(m,n')}dx\\ &\leq \int_{0}^1 \left| \sum_m \sum_n \sum_{n'}e(nx) \overline{e(n'x)}\phi(m,n) \overline{\phi(m,n')}\right|dx\\ &\leq \Delta \sum_{n} 1, \end{align*} by the hypothesis, but this obviously won't work. Here $e(x) := e^{2\pi i x}$ (thus $|e(x)| = 1$ for any real $x$). Is there a trick here that may work? Or perhaps the claim is not true? Thanks and sorry for the long post!
Eventually I realized (what should have been obvious) that in one of the steps we could apply the original duality lemma for quadratic forms, which holds in our case by our hypothesis. Thus the bilinear case turns out to be a corollary of the lemma for quadratic forms. I give the proof and make it available. Hope it's correct!
Proposition: Let $\phi : \mathbb{Z}^2 \to \mathbb{C}$ and suppose there is a constant $\Delta$ such that for all complex numbers $\gamma_n$ we have \begin{equation}\label{eqn: A} \sum_m \left| \sum_n \gamma_n \phi(m,n)\right|^2 \leq \Delta \left\| \gamma \right\|^2. \end{equation} Then for all complex numbers $\alpha_m, \beta_m$ we have $$ \left|\sum_n \sum_m \sum_{m'} \alpha_m \overline{\beta_{m'}} \phi(m,n) \overline{\phi(m',n)}\right| \leq \Delta \left\| \alpha \right\|\left\| \beta \right\|. $$ Here $m,m'$ run over some set $M$ and $n,n'$ over some set $N$.
Proof: Let $\alpha_m, \beta_m$ be arbitrary complex numbers. Then by assumption and the duality lemma for quadratic forms we have in particular that $$ \sum_n \left| \sum_m \beta_m \phi(m,n) \right|^2 \leq \Delta \left\| \beta\right\|^2. $$ Let $\gamma_n := \sum_{m} \overline{\beta_m \phi(m,n)}$. The assumption implies $$ \sum_m \left| \sum_n \gamma_n \phi(m,n)\right|^2 \leq \Delta \left\| \gamma \right\|^2. $$ Moreover $$ \left\| \gamma \right\|^2 = \sum_n |\gamma_n|^2 = \sum_n\left| \sum_{m} \overline{\beta_m \phi(m,n)}\right|^2 = \sum_n\left| \sum_{m} \beta_m \phi(m,n)\right|^2 \leq \Delta \left\| \beta\right\|^2. $$ Now rearranging terms and applying the Cauchy-Schwarz inequality yields, together with the previous facts, \begin{align*} & \left|\sum_n \sum_m \sum_{m'} \alpha_m \overline{\beta_{m'}} \phi(m,n) \overline{\phi(m',n)}\right| = \left| \sum_m \alpha_m \sum_n \sum_{m'}\overline{\beta_{m'}\phi(m',n)} \phi(m,n)\right|\\ &= \left| \sum_m \alpha_m \sum_n \gamma_n \phi(m,n)\right| \leq \left( \sum_m |\alpha_m|^2 \right)^{1/2} \left( \sum_m \left| \sum_n \gamma_n \phi(m,n) \right|^2 \right)^{1/2}\\ &\leq \Delta^{1/2} \left\| \alpha \right\| \left\| \gamma \right\|\\ & \leq \Delta \left\| \alpha \right\|\left\| \beta\right\|, \end{align*} as required.