Bellman-like integral inequality

Question

I have a weakly decreasing, continuous function $w(\cdot)\geq 0$ on $[0,T]$, with $w(0)=1$. I also have a continuous, decreasing function $b(\cdot)$, where $b(t)\geq 0$ for $t\in[0,T]$. I know that \begin{align*} w(t)\leq 1-\int_0^tw(s)b(s)ds \end{align*} for all $t\in[0,T]$. I am trying to determine the maximum possible value $w(T)$ under these conditions. Does anyone have suggestions besides trying to formulate this as a control problem? Are there functional/integral inequalities that could work? The ones I have seen all assume that the integral enters with a positive sign on the right hand side.

Answer 1

The optimal value of $w(T)$ is $$ w^* = \left(1+ \int_0^T b(t) \mathrm{d} t\right)^{-1}. $$

Upper bound. Since $w$ is non-increasing, $\int_0^T w b \geq w(T) \int_0^T b$. Thus, your constraint applied at the final time is $w(T) \leq 1- w(T) \int_0^T b$, which yields $w(T) \leq w^*$.

Lower bound. Obviously, as soon as $b \neq 0$, this optimal value is not achieved by any continuous $w$ because such a $w$ would satisfy both your initial condition $w(0) = 1$ and the condition $w(t) \equiv w^* < 1$ for all $t \in [0,T]$ to achieve the equality $\int_0^T w b = w^* \int_0^T b$ in the upper bound above.

Nevertheless, you can construct a sequence of functions for which $w(T) \to w^*$, for example by taking $w_n$ defined as follows: $w_n(t) = w^* - \frac{1}{n}$ for $t \geq t_n$ and $w_n$ decreasing linearly from $1$ to $w_n(t_n)$ on $[0,t_n]$, where $0 < t_n \ll 1$ is small enough so that the constraint is satisfied.