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Denoting $e(x)$ for $e^{2i\pi x}$, set

$$E(R):=\left\{f\ \left|\ f(x)=\sum_{r=0}^{R-1}a_re(rx)\mbox{ where }a_r\in\mathbb{C}\ \forall r\mbox{ and }\sum_r|a_r|^2=1\right\}\right.$$

$h(a,R):=\inf_{f\in E(R)}\int_0^a|f(x)|^2dx$ with $0\leq a\leq 1$ .

I am curious to know the behavior of $h(a,R)$ particularly if $R\rightarrow\infty$

Thank you in advance for an idea!

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This is directly related to the so-called prolate matrix (see, for example, this paper), and specifically, its minimal eigenvalue. To see this, note that $$ \int^a_0 | f(x) |^2 dx = \sum^{R-1}_{r,s=0} a_r \overline{a_s} \int^{a}_{0} e^{2 \pi i (r-s) x} d x. $$ Let $b_r = e^{\pi i r a} a_r$, and write $A = \{ A_{r-s} \}^{R-1}_{r,s=0}$ for the prolate matrix, i.e. $$ A_0 = a,\qquad A_r = \frac{\sin \pi r a}{\pi r},\ r \neq 0. $$ If $b=(b_0,\ldots,b_{R-1})$ then $$ h(a,R) = \min_{\sum_r |b_r|^2 = 1} b^* A b = \lambda_{\min}(A). $$ Eqn (3.2) of the above paper now gives the exact formula for $h(a,R)$.

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