Minimum infinity norm control problem I am having trouble understanding Example 2 of section 5.9 of Luenberger's Optimization by Vector Space Methods. The problem is to select a current $u(t)$ on $[0,1]$ to drive a motor governed by
$$ \ddot{\theta}(t) + \dot{\theta}(t) = u(t) $$
from $\theta(0) = \dot{\theta}(0) = 0$ to $\theta(1) = 1, \dot{\theta}(1) = 0$ minimizing $\max_{0 \leq t \leq 1} |u(t)|$. Integrating the differential equation allows one to express the constraints as
$$
\dot{\theta}(1) = \int_0^1{e^{t-1}u(t)dt} = 0 \\
\theta(1) = \int_0^1{(u(t) - \ddot{\theta}(t))dt} = \int_0^1{(1 - e^{t-1})u(t)dt} = 1
$$
Defining $y_1$, $y_2 \in L_1[0,1]$ by $y_1(t) = e^{t-1}$, $y_2(t) = 1 - e^{t-1}$, we seek $u \in L_{\infty}[0,1]$ of minimum norm satisfying
$$
\langle y_1, u \rangle = 0 \\
\langle y_2, u \rangle = 1,
$$
where $\langle x, f \rangle$ denotes $f(x)$, $f$ being a bounded linear functional.
By a previous theorem, $\min ||u|| = \max_{||a_1 y_1 + a_2 y_2|| \leq 1} a_2$. Luenberger writes "Maximization of $a_2$ subject to this constraint is a straightforward task, but we do not carry out the necessary computations.". He then shows that the optimal $u$ is "bang-bang". I understand the transformation of the problem and why $u$ is bang-bang, and I know that once I have $a_1$ and $a_2$ I may determine $u$, since it changes sign at the same time as $a_1 y_1 + a_2 y_2$, and its absolute value is equal to $a_2$. But I do not know how to determine $a_1$ and $a_2$, and I do not see how maximization of $a_2$ subject to the integral constraint is a straightforward task. Any hints would be very welcome.
 A: I was able to solve this with some outside help. Since there seemed to be some interest in the problem (judging by the upvotes), I am posting the solution here.
We consider three separate cases:
$$
(a_1 - a_2)\mathrm{e}^{t-1} + a_2 > 0 \text{ for all } t \in (0, 1) \\
(a_1 - a_2)\mathrm{e}^{t-1} + a_2 < 0 \text{ for all } t \in (0, 1) \\
(a_1 - a_2)\mathrm{e}^{t-1} + a_2 = 0 \text{ for some } t \in (0, 1)
$$
In this way we can calculate the restriction integral directly and obtain simpler restrictions in order to apply standard methods.
In the first case,
$$
\int_0^1{((a_1 - a_2)\mathrm{e}^{t-1} + a_2)\mathrm{d}t} \leq 1 \iff a_2 \leq \mathrm{e} - (\mathrm{e} - 1)a_1
$$
Also, since the function is nonnegative at the endpoints,
$$
a_1 \geq 0
$$
and
$$
a_2 \geq -\frac{1}{\mathrm{e} - 1}a_1
$$
Maximization of $a_2$ can be performed geometrically; the optimum is found to be $(a_1, a_2) = (0, \mathrm{e})$.
The second case is similarly handled, the solution is $(a_1 ,a_2) = (\frac{\mathrm{e}(1 - \mathrm{e})}{(\mathrm{e} - 1)^{2} - 1}, \frac{\mathrm{e}}{(\mathrm{e} - 1)^{2} - 1})$. This is worse than the solution for the first case, so we discard it.
The third case is a little harder. If $(a_1, a_2)$ is feasible, so is $(-a_1, -a_2)$, hence since we are maximizing $a_2$ we may assume $a_2 \geq 0$. If $a_2 = 0$, we get $u(t) = 0$ almost everywhere, which is not a solution (by the second restriction). Therefore $a_2 > 0$. If $(a_1 - a_2)\mathrm{e}^{t_0-1} + a_2 = 0$, then $t_0 = 1 + \log{(a_2/(a_2 - a_1))}$. Thus we need to have $a_2 - a_1 > 0$, which means $(a_1 - a_2)\mathrm{e}^{t-1} + a_2$ is decreasing in $t$, so it will be postive first and then negative. Now we can integrate the constraint:
$$
\int_0^1{|(a_1 - a_2)\mathrm{e}^{t-1} + a_2| \mathrm{d}t} \leq 1 \iff \\
\int_0^{t_0}{((a_1 - a_2)\mathrm{e}^{t-1} + a_2)\mathrm{d}t} + \int_{t_0}^1{((a_2 - a_1)\mathrm{e}^{t-1} - a_2)\mathrm{d}t} \leq 1 \iff \\
\mathrm{e}^{-1}a_2 - (1 + \mathrm{e}^{-1})a_1 + 2a_2\log{\frac{a_2}{a_2-a_1}} \leq 1.
$$
If we can show that the optimum is achieved with equality in this restriction, this will be reduced to an equality and then we can apply the Lagrange multiplier method. Let $(a_1, a_2)$ be feasible with
$$
\int_0^1{|(a_1 - a_2)\mathrm{e}^{t-1} + a_2| \mathrm{d}t} = 1 - \delta
$$
for some $\delta \in (0, 1)$.
Then
$$
\int_0^1{|(a_1 - (a_2 + \mathrm{e}\delta)\mathrm{e}^{t-1} + a_2 + \mathrm{e}\delta| \mathrm{d}t} = \int_0^1{|(a_1 - a_2)\mathrm{e}^{t-1} + a_2 + \mathrm{e}\delta(1 - \mathrm{e}^{t-1})| \mathrm{d}t} \\
\leq \int_0^1{|(a_1 - a_2)\mathrm{e}^{t-1} + a_2| \mathrm{d}t} + \mathrm{e}\delta\int_0^1{|(1 - \mathrm{e}^{t-1})| \mathrm{d}t} \\
= 1 - \delta + \mathrm{e}\delta(1 - (1 - \mathrm{e}^{-1})) \\
= 1
$$
In other words, $(a_1, a_2 + \mathrm{e}\delta)$ is feasible. So $(a_1, a_2)$ cannot be optimal. Hence the optimal point must be achieved with equality in the restriction.
Now we form the Lagrangian $L(a_1, a_2, \lambda) = a_2 + \lambda (\mathrm{e}^{-1}a_2 - (1 + \mathrm{e}^{-1})a_1 + 2a_2\log{\frac{a_2}{a_2-a_1}} - 1)$, and apply the Lagrange multiplier method. This gives a simple system of equations with solution $((1 - \frac{2}{1 + \mathrm{e}^{-1}})\frac{1}{\log{\frac{(\mathrm{e}+1)^2}{4\mathrm{e}}}}, \frac{1}{\log{\frac{(\mathrm{e}+1)^2}{4\mathrm{e}}}})$. This has a greater value of $a_2$ than the solution in the first case, so it must be the solution.
