  One possibility of defining a quantum R\'enyi $\alpha$-divergence of two
quantum states is to optimize the classical R\'enyi $\alpha$-divergence of
their post-measurement probability distributions over all possible measurements
(measured R\'enyi divergence), and maybe regularize these quantities over
multiple copies of the two states (regularized measured R\'enyi
$\alpha$-divergence). A key observation behind the theorem for the strong
converse exponent of asymptotic binary quantum state discrimination is that the
regularized measured R\'enyi $\alpha$-divergence coincides with the sandwiched
R\'enyi $\alpha$-divergence when $\alpha>1$. Moreover, it also follows from the
same theorem that to achieve this, it is sufficient to consider $2$-outcome
measurements (tests) for any number of copies (this is somewhat surprising, as
achieving the measured R\'enyi $\alpha$-divergence for $n$ copies might require
a number of measurement outcomes that diverges in $n$, in general). In view of
this, it seems natural to expect the same when $\alpha<1$; however, we show
that this is not the case. In fact, we show that even for commuting states
(classical case) the regularized quantity attainable using $2$-outcome
measurements is in general strictly smaller than the R\'enyi
$\alpha$-divergence (which is unique in the classical case). In the general
quantum case this shows that the above "regularized test-measured" R\'enyi
$\alpha$-divergence is not even a quantum extension of the classical R\'enyi
divergence when $\alpha<1$, in sharp contrast to the $\alpha>1$ case.
