Under which condition does the heuristic find the true global minimum? Consider the unconstrained programming problem with a high-dimensional, smooth function $f(x_1,x_2,\ldots x_N)$. 
Because $N$ is large, such kind of heuristics is sometimes used:
-- Fix $x_2…x_N$ to 0, and minimize $x_1\rightarrow f(x1, 0, 0,..0)$. Suppose this step happens to get the true minimum point $x_1^*$.
-- Fix $x_3,….X_N$ to 0, and minimize $x_2\rightarrow f(x_1^*, x_2, 0,..0)$. Suppose this step happens to get the true minimum point $x_2^*$.
-- Eventually, after $x_1^*, ….x_{N-1}^*$ are calculated, we can get $x_N^*$. 
My question is: under which  condition does $(x_1^*,x_2^*…x_N^*)$ coincide with a global minimum point of $f$?
 A: The simplest case that I can think of is:
$$
f(X) = \sum_ig_i(x_i)
$$
where $X=\{x_1, x_2, ..., x_N\}$. That is, the function $f(X)$ is a sum of independent functions of the $x_i$'s. You can see intuitively why this meets your criteria, because the only way for the sum to be minimal is if all the individual terms are also at their minimum. If one term could be reduced further, then that would decrease the sum, so we wouldn't be at the minimum. Thus, minimizing the composite function is equivalent to minimizing the individual components separately.
Formally, you can work it out by noting that at the minimum of a function, its derivative is equal to 0, and the derivative of a sum of functions is equal to the sum of the derivatives, i.e.:
$$
\frac{\delta f(X)}{\delta x_i}=\sum_j\frac{\delta g_j(x_j)}{\delta x_i}
$$
Since only the $i$-th function actually depends on $x_i$, all derivatives where $j\neq i$ are 0, such that the expression simplifies to:
$$
\frac{\delta f(X)}{\delta x_i}=\frac{d g_i(x_i)}{d x_i}
$$
and thus:
$$
\frac{\delta f(X)}{\delta x_i}=0 \iff \frac{d g_i(x_i)}{d x_i}=0
$$
which means that the derivative of $f(X)$ is 0 if and only if we find the $x_i$'s for which the derivatives of each of the component functions $g_i(x_i)$ are 0, which we can do separately for each component function because each only depends on its own $x_i$.
Note that I'm assuming here that the component functions $g_i(x_i)$ have a single unique minimum. If one of the components has multiple local minima, then those are also local minima of $f(X)$, and so the solution the algorithm finds may be not be the global minimum.
