Fix a positive constant $q=O(1)$ (say $1.5$). I am trying to find a function $\ell(x):[0, 1] \to \mathbb{R}_{\geq 0}$ that satisfies $\int_0^1 \ell(x) dx \leq q$ and minimizes the expression

$$\int_0^1 e^{-qx}\ell(x) dx - (1-e^{-q})(1-e^{-\int_0^1 \ell(x)})$$

My colleague told me that variational calculus might be useful here, but looking at the Wikipedia page, it seems to apply to functions of the form $\int_{x_1}^{x_2}L(x, y(x), y'(x)) dx$ for some function $L$. However, I can't write my specific problem in that format. Is there a general method to approach these optimization problems (I just need a pointer to read more, no need to even solve this specific problem!)

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    $\begingroup$ Once you know what value $a$ you want to give to $a=\int_0^1 \ell$, the optimal configuration would obviously be $\ell=a\delta_1$. If you allow this, then finding the optimal value of $a$ is now a simple calculus problem, or if not, then you don't have a minimizer, but can get arbitrarily close to this situation. $\endgroup$ Apr 2 at 23:58

1 Answer 1


Assuming that $q > 1$ and letting $\lambda = 1 - e^{-q}$, we can rewrite the problem as follows: $$ \int_0^1 e^{-qx} \ell(x) dx + \lambda \exp \left( -\int_0^1 \ell(x) dx \right) \to \min_\ell. $$ It is clear that minimizing the first term means requires $\ell$ to have more mass near $1$, whereas minimization of the second in indifferent to the distribution of mass as long as the total mass remains the same. Hence, we can push all the mass of $\ell$ to $1$, which means the minimizing $\ell$ must not be a function, but a measure $m \delta_1$, where $m$ is the total measure. Plugging this into the functional, we obtain a new optimization problem: $$ e^{-q} m + \lambda e^{-m} \to \min_{m}. $$ Solving this, gives $$ m = q + \ln \lambda. $$ Therefore, $$ \ell = \left( q + \ln \left( 1 - e^{-q} \right) \right) \delta_1. $$


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