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I am trying to solve an optimization problem of this form with Calculus of variations: $$ minimize \space J[f(t)]=\int_a^bf(t)\cdot g(t)dt $$

$$ subject \space to \int_a^bf(t)dt=K, $$ which is written as: (using Lagrange multipliers) $$ J[f(t),\lambda]=\int_a^bf(t)\cdot g(t)- \lambda \cdot (f(t)-\frac {K}{(b-a)})dt. $$ But when I use Euler-Lagrange equation, both $f(t)$s are omitted and I have only: $$ g(t)- \lambda =0 $$ I know that $g(t)$ is an increasing function and so $f(t)$ should be decreasing to minimize the integral. But I want to find the optimal answer. Can anybody help me on solving this problem?

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    $f$ and $g$ are really independent? if so then the solution is exactly what you have obtained, i tried to solve the problem without using $E-L$ equations and i've come up with the same result. – 2017-01-14
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    @Francesco Alem Thanks for you answer, but how can I use this answer to find $f(t)$. I am interested in finding the $f(t)$ function and I know that the answer is dependent on it, because $g(t)$ is an increasing function and so $f(t)$ should be decreasing to minimize the integral. – 2017-01-15
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    based on your observation i would say that the optimal solution is to be written in terms of dirac delta function: $$ f(t)=K \delta (t-a) $$ which satisfies the constraint. and for which the functional $$ J\[f\]=Kg(a) $$ – 2017-01-15
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    @Francesco Alem. Thanks a lot – 2017-01-17

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Hints:

  1. In this answer, we assume that the given function $g$ is real Lebesgue measurable function and not constant almost everywhere (Else OP's constrained functional $J[f]$ is independent of allowed $f$). With this assumption, there are no extremal $f$-functions.

  2. In more detail, construct two Lebesgue measurable sets $A_i\subseteq [a,b]$ such that $$ \det\begin{pmatrix} r_1 & m(A_1) \cr r_2 & m(A_2) \end{pmatrix}~\neq~ ~0,\tag{1}$$ where $$ r_i~:=~\int \! dm ~1_{A_i} ~g, \qquad i\in\{1,2\}.\tag{2}$$ Let $$f ~=~\sum_{i=1}^2 c_i 1_{A_i} , \qquad c_i~\in~\mathbb{R}.\tag{3} $$ Then $$J[f]~=~\sum_{i=1}^2 r_ic_i , \qquad K~=~\int \! dm ~f~=~ \sum_{i=1}^2 m(A_i)c_i. \tag{4} $$ Since the discriminant (1) is non-zero, the two linear equations (4) can be solved for the two unknowns $c_1$ and $c_2$ for any value of OP's functional. Hence there cannot be any extremal $f$-function.