I am curious about the general relation between *nested, successive minimization* (M1) and *asymptotic minimization* (M2) as defined in the following. What one wants is to implicitly minimize a sequence of functions where each respective domain is based on the priorly obtained minimizers.

**Problem setting.** Let $D \subset \mathbb{R}^N$ be some suitable domain and let
$$g_k: D \rightarrow \mathbb{R}, \quad k = 0,\ldots,n$$
be continuous functions that have well defined minima. Further, let
$$G_\gamma: D \rightarrow \mathbb{R}, \quad G_\gamma(a) := \sum_{k = 0}^n \gamma^{n-k} g_k(a), \quad \gamma > 0.$$

**(M1)**: Let $A_{n+1} := D$ and successively define
$$A_k := \{ a \in A_{k+1} \mid g_k(a) = \min_{b \in A_{k+1}} g_k(b) \}, \quad k = n,n-1,\ldots,0.$$
Then set $S_{M1} := A_0$.

**(M2)**: Define
$$ S_{M2} := \{ a^\ast \in D \mid \exists (a_\gamma)_{\gamma > 0} \subset D, \ a^\ast = \lim_{\gamma \searrow 0} a_\gamma, \ G_\gamma(a_\gamma) = \min_{a \in D} G_\gamma(a)\}.$$
So this set consists of the limits of minimizers of $G_\gamma$ for $\gamma \searrow 0$.

**Question.**
How little assumptions are required such that $S_{M2} = S_{M1}$ or at least $S_{M2} \subset S_{M1}$?

The interesting situation is of course when the functions $g_k$ are flat around their minimizers.

I am fairly convinced that the sets are commonly equal. In other words, it requires *some work* to come up with functions for which the sets are not equal, but this is if course a completely informal statement.

Despite some knowledge about the situation (for $n > 1$), I struggle to find literature on this, basically because I do not really know where to start.

- Can you provide literature about this problems, or suggest under which phrase to search?
- Can you provide less restrictive assumptions as (A1) laid out below?
- Can you provide further reaching counter examples as (C1) below?
- Does the relation always hold if one considers
*local*instead of global minima?

**(A1)** Let $D = \mathbb{R}^N$ and $s \geq 0$. Assume that for $k = s+1,\ldots,n$
$$\underset{a \in D}{\mathrm{argmin}} \ g_k(a) \subset \underset{a \in D}{\mathrm{argmin}} \ g_{k+1}(a).$$
Then $S_{M2} \subset A_s$. Thus, if additionally $A_s$ contains only one element and $S_{M2}$ is not empty, then $S_{M2} = S_{M1}$.

**(C1)**: For $n = 2$, let
$$g_2(x) := \begin{cases}
0 &,\ x < 0 \\
x^2 &,\ x \geq 0
\end{cases}, \quad
g_1(x) := \begin{cases}
0 &,\ x < 0 \\
-x &,\ x \in [0,1] \\
-1 &,\ x > 1
\end{cases}, \quad
g_0(x) := \begin{cases}
\frac{2|x + \frac{1}{2}| - 1}{8} &,\ x < 0 \\
0 &,\ x \geq 0
\end{cases}.
$$
Then $A_2 = (-\infty,0] = A_1$ and thus $S_{M2} = A_0 = \{-\frac{1}{2}\}$. On the other hand, it is
$$G_\gamma(x) = \begin{cases}
\frac{2|x + \frac{1}{2}| - 1}{8} \gamma^2 &,\ x < 0 \\
x^2 -\gamma x &,\ x \in [0,1] \\
x^2 + \gamma x - 2\gamma &,\ x > 1
\end{cases}$$
Short calculation shows $-\frac{\gamma^2}{8} = G_\gamma(-\frac{1}{2}) = \min_{x < 0} G_\gamma(x)$ and $-\frac{\gamma^2}{4} = G_\gamma(\frac{\gamma}{2}) = \min_{x \geq 0} G_\gamma(x)$. So the minimum within $(-\infty,0]$ is *hidden*, and we obtain $S_{M2} = \{0\}$.