Minimization is equal to minimization over intervals

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Given a function class $\mathcal{F}$ of square integrable functions and a function $g$, let

$$ f^* := \underset{f \in \mathcal{F}}{\operatorname{argmin}} \int_{0}^{\infty} f(x) (f(x) - g(x)) {\rm d} x $$

Is it possible to split the minimization problem as follows?

$$ \underset{f \in \mathcal{F}}{\operatorname{argmin}} \int_{0}^{a} f(x) (f(x) - g(x)) {\rm d} x + \underset{f \in \mathcal{F}}{\operatorname{argmin}} \int_{a}^{\infty} f(x) (f(x) - g(x)) {\rm d} x $$

The idea is that the two argmin should optimize on separate intervals so it should still give $f^*$, but I couldn't formalize it.

E.g., the class $\mathcal{F}$ is the class of all decreasing square integrable bounded and right continuous functions on $[0,\infty)$.