# Minimizing expressions involving function subject to integral constraint

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!)

• 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. Apr 2 at 23:58

## 1 Answer

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, $$\ell = \left( q + \ln \left( 1 - e^{-q} \right) \right) \delta_1.$$