In the special case when $r=1$, Montgomery and Odlyzko [3] observed that this optimization problem had already been solved using prolate spheroidal wave functions (see comments below). Here is how they reduced the problem to one that was already solved. In this case, we have $$ M(c) = c+ \int_0^1\int_0^1 f(u) f(v) \frac{\sin(\pi c(u-v))}{\pi(u-v)} \ dv \ du. $$ The double integral on the right-hand side is $$ \int_{-c/2}^{c/2} \left| \int_0^1 f(v) e^{2\pi i v} dv \right|^2 dt :=I(c),$$$$ \int_{-c/2}^{c/2} \left| \int_0^1 f(v) e^{2\pi i t v} dv \right|^2 dt :=I(c),$$ say. They then observed that choosing $$ f(x) = aR_{00}^{(1)}[\pi c/2,2x-1]$$ maximizes $I(c)$, where $R_{mn}^{(1)}[c,x]$ is the radial prolate spheroidal wave function of the first kind of order m and degree n, and a is a constant to be chosen according to our above normalization.
Theorem: Let $\alpha(c)$ be the least number such that $$ \int_{-c/2}^{c/2} \left|\int_0^1 f(x) e^{2\pi i x} dx \right|^2 dt \le \alpha(c) \int_0^1 |f(x)|^2 dx $$$$ \int_{-c/2}^{c/2} \left|\int_0^1 f(x) e^{2\pi i t x} dx \right|^2 dt \le \alpha(c) \int_0^1 |f(x)|^2 dx $$ for all $f\in L^2[0,1]$. Then $\alpha(c)$ is strictly increasing, $\alpha(c)\lt c$ for $c \gt 0$, $\alpha(c)\sim c$ as $c\to 0^+$, and $\alpha(c)\to 1$ as $c\to \infty$. Moreover, equality is achieved in the above inequality if $f(x) = R_{00}^{(1)}[\pi c/2,2x-1]$.