The order of derivatives implies reversed order of minimizers

Question

Suppose that $f_1$ and $f_2$ are both differentiable functions defined on $\mathbb{R}$ that have finite minimizers, $S_1$ and $S_2$, respectively, and that $f_1' \le f_2'$. Show that if $S_1$ and $S_2$ are unique minimizers, then $S_1 \ge S_2$.

I understand the "minimizers" to mean global minimizers. I tried showing it by contradiction: Suppose $S_1 < S_2$. Since $f_1(S_1)<f_1(S_2)$, $f_2(S_1)>f_2(S_2)$, if $f_1'$ and $f_2'$ are integrable, then $$0>f_2(S_2)-f_2(S_1) = \int_{S_1}^{S_2}f_2'(x)\textrm{d}x \ge \int_{S_1}^{S_2}f_1'(x)\textrm{d}x = f_1(S_2)-f_1(S_1)$$ contradicting that $f_1(S_1)<f_1(S_2)$. However, I don't know how to do the trick when the derivatives are not necessarily integrable. Could anyone please help me with this problem?

This problem is from Evan L. Porteus, Foundations of Stochastic Inventory Theory, p. 115.

Answer 1

The problem becomes more transparent if we write $g=f_2-f_1$ and restate the claim as

adding a nondecreasing function (namely, $g$) to $f_1$ cannot move the minimizer to the right.

Which is intuitively obvious, and easy to prove. We already have $f_1(x)>f_1(S_1)$ for all $x>S_1$. Adding $g$ preserves this: $$f_1(x)+g(x)>f_1(S_1)+g(S_1)\quad \text{ for all }x>S_1$$