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Technically, I am working on an optimal control problem. However, through some trickery, I managed to eliminate the dynamics. What I am left with is the following minimization problem:

$$ \min_{g \in L^1} \int_0^\infty \big[\alpha(sm) - g \left( s \cdot \|g\|_1 \right) \big]^2 {\rm d} s $$

Here, $\|g\|_1 := \int_0^\infty |g(s)| ds$, and $m$ is simply a constant. Further, $\alpha$ is a function that can be assumed to be arbitrarily smooth, though minimally $L^1 = \{f : \|f\|_1 <\infty \}$. I have the following questions:

  1. How do you solve this problem?

  2. Does this problem fall into a class of larger problems that have a name (Quite frankly I have never seen this implicit dependence before)?

  3. Is there literature that discusses these kinds of problems?

  4. How hard is this equation to solve numerically? Just a ballpark, like something between "takes a minute on your local laptop" and "Ask your university for more compute".

  5. Naturally one would hope that the equation $$ \alpha(sm) = g(s\cdot \|g\|) $$ is solvable for $g$. In my mind this can not be done for general $\alpha \in L^1$. However, if somebody can proof me wrong here that would also be useful, since it means that I did something significantly wrong in my derivation.

With "these kinds of problems" I always refer to minimization problems over function spaces that also have the implicit relation $g(s \cdot \|g\|)$.

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    $\begingroup$ It would be nice to know what your original optimal control problem is $\endgroup$ Commented Jun 3 at 11:29
  • $\begingroup$ Note that for arbitrary $g \in \mathrm{L}^{1}$ and arbitrary $m > 0$ you can define $\alpha(s) = g(s\tfrac{\|{g}\|}{m})$. Hence, there are plenty of $\alpha$, where you can find a $g$. Of course this is reversed direction, but still it shows that sometimes you can solve $\alpha(sm) = g(s\|{g}\|)$. $\endgroup$ Commented Jun 3 at 12:58
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    $\begingroup$ Now that I think about it: In fact you can solve $\alpha(sm) = g(s\|g\|)$ if and only if $m = \|a\|$. This is direct consequence of $\|g(\cdot \|g\|)\| = 1$. $\endgroup$ Commented Jun 3 at 13:10
  • $\begingroup$ @NathanaelSkrepek Ah nice thank you that helps! Also it fits reasonably well with my intuition of the whole problem. $\endgroup$ Commented Jun 3 at 13:28

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