An optimization problem in $L^1(0,1)$ Is there any non-negative function $f(t)$ that minimizes $\int_0^1e^{\int_0^tf(s)ds}dt$ and satisfies $\int_0^1sf(s)ds =1$?
I guess there is not, because the exponential is minimized if $\int_0^tf(s)ds$ is as small as possible. However, for such functions there can never be $\int_0^1sf(s)ds\neq 0$.
 A: There is no real-valued function that is a minimizer.  Specifically, consider searching for real-valued functions $f:[0,1]\rightarrow [0, \infty)$ to solve
\begin{align}
\mbox{Minimize:} \quad & \int_0^1 e^{\int_0^t f(s)ds}dt\\
\mbox{Subject to:} \quad & \int_0^1 sf(s)ds = 1
\end{align}
Define $v^*$ as the infimum objective value over all functions that satisfy the constraints. It is clear that all nonnegative functions $f(t)$ yield an objective value $\int_0^1 e^{\int_0^t f(s)ds}dt\geq 1$ and so $v^*\geq 1$. 

Fix $z$ such that $0<z<1$ and define the nonnegative real-valued function
$$ f_z(t) = \left\{ \begin{array}{ll}
\frac{2}{1-z^2} &\mbox{ if $z<t<1$} \\
0  & \mbox{ otherwise} 
\end{array}
\right.$$
This has the following properties: 


*

*$\int_0^1 sf_z(s)ds = 1$ 

*$1\leq \int_0^1 e^{\int_0^tf_z(s)ds}dt \leq z + (1-z)e^{\frac{2}{1+z}}$ [see derivation below]
Notice that $\lim_{z\rightarrow 1} \int_0^1e^{\int_0^t f_z(s)ds}dt = 1$.  So the infimum value of the objective function, over all functions that satisfy the constraints, is 1. That is, $v^*=1$.  But this infimum cannot be achieved because it would require $f(t)=0$ for almost all $t \in [0,1]$, which would mean $\int_0^1sf(s)ds=0$, which does not satisfy the constraint. 

A "generalized" function that is a minimizer is: 
$$ f(t) = \delta(t-1)$$
where $\delta(t)$ is the unit impulse function.  Of course, this is not a real-valued function.  Notice that the $f_z(t)$ functions can be viewed as approximations of $\delta(t-1)$ when the value $z$ is pushed very close to $1$.

To prove the above inequality regarding $f_z(t)$ we have 
\begin{align}
\int_0^1 e^{\int_0^t f_z(s)ds} &= \int_0^z e^0dt + \int_z^1 e^{\int_0^t f_z(s)ds}dt \\
&\leq z + \int_z^1 e^{\int_0^1f_z(s)ds}dt\\
&= z + (1-z)e^{\frac{2}{1+z}}
\end{align}
