$\newcommand{\D}{\mathbb{D}}$ Let $H^\infty(\D)$ be the space of bounded analytic functions in the unit disc $\D$. For a function $f(z) = \sum_{n=0}^\infty a_nz^n$ we can define its truncation as $$T_xf(z) = \sum_{n\le x}a_nz^n.$$
$T_x$ is a bounded linear operator on $H^\infty (\D)$ and it is known that $||T_x||\asymp \log x$ as $x\to \infty$ (although I don't know the precise constant in this equivalence).
Similarly, let $H^\infty(\D^2)$ denote the space of bounded analytic functions of two complex variables in the unit disc. Given numbers $a, b \ge 1$ and $x>0$, we can define an operator which acts on $f(z, w) = \sum_{n = 0}^\infty \sum_{m = 0}^\infty a_{n, m}z^nw^m$ as $$T_{a, b, x}f(z, w) = \sum_{an+bm \le x}a_{n, m}z^nw^m.$$
It is a "truncation" operator which leaves only terms lying in the triangle with vertices $(0, 0), (x/a, 0)$ and $(0, x/b)$.
I can prove that $||T_{a, b, x}||\lesssim \log x$ with constant which doesn't depend on $a$ and $b$ (something like $20\log x$ or so). But what I want to know is can we get any advantage at all by using extra dimension? More precisely
Question. Do there exist numbers $a, b\ge 1$ and $x > 0$ such that $||T_{a, b, x}|| > ||T_x||$?
Negative answer to (a higher-dimensional version of) this question would resolve a certain open problem about spaces of Dirichlet series in a very unexpected way but I can not imagine this to be false so I'm asking it here.
Note that conditions $a, b\ge 1$ are clearly necessary since otherwise $T_{a, b, x}$ may contain $T_y$ for some $y > x$ as a suboperator and the answer would be trivially yes. On the other hand, for some $a, b$ it is not hard to show that this could never happened. For example, if $a, b\in \mathbb{N}$ then I can prove that $||T_{a, b, x}||\le ||T_x||$.
