Problem
Consider two d x d complex matrices, R and S, whose entries lie in the unit disk:
$\quad |R_{i,j}|<1 \quad$ and $\quad |S_{i,j}|<1 $.
Say that R is constructed by randomly choosing complex numbers from the unit disk, but S is constructed as
$\quad S_{i,j} = f(i/d,j/d)$
where $f(x,y)$ is a smooth function for $x,y \in [0,1]$, with $|f(x,y)|<1$. In other words, the entries of S are smooth functions of the indices (in the limit of large d), but those of R are not.
Question
How do the trace norms
$\quad ||R||=Tr[\sqrt{R^\dagger R}] \quad$ and $\quad ||S||=Tr[\sqrt{S^\dagger S}]$
of these matrices behave as $d \to \infty$ ?
Numerical Evidence
A few lines of Mathematica strongly suggest that
$\quad ||R|| \propto d^{3/2}$
but
$\quad ||S|| \propto d$
for large d. (The proportionality constants depend on the probability distribution used to pick numbers from the unit disk for R and the function $f(x,y)$ used to pick entries for S, respectively.)
What explains this behavior?
Addendum
After Willie's excellent answer below, I thought I'd mention that it's really fast to see the scaling behavior once you discretize the function. Let $F$ be some matrix of discrete values for the function, and let $J_n$ be the $n \times n$ matrix with all elements equal to unity.
$||F \otimes J_n|| = \mathrm{Tr} \sqrt {(F^\dagger \otimes J_n)( F \otimes J_n)} = \mathrm{Tr} \sqrt {(F^\dagger F) \otimes (J_n J_n)} = ||F|| \cdot ||J_n|| = n ||F||$
Basically, the idea is that once the dimension of $F$ is large enough to capture the important detail in the function, increasing the dimension is really just increasing the dimension of $J$.
