# Proof that maximizing a function is equivalent to minimizing its negative

The statement that maximizing a function over its argument is equivalent to minimizing that function over the same argument with a sign change seems to be accepted as trivial wherever I look (MSE, proofwiki, internet search, textbooks outside of optimization theory). Intuitively, if you have some function of a single variable that has a global maximum, and you "flip it over" by changing the sign, the global maximum is now a global minimum. I can appreciate that.

However, it seems to me that math is all about meticulous examination of surprising subtleties. Does anyone know of a good way to prove this statement? I just don't feel comfortable without it.

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$f$ is maximized at $x$ if $f(x)\ge f(y)$ for all $y$. $-f$ is minimized at $x$ if $-f(x)\le -f(y)$ for all $y$. Can you complete the proof? – Rahul Mar 18 '13 at 9:03
So there's no tricks, no nothing about the spaces the function is mapping, no pathological anything? That's wierd! – Trevor Alexander Mar 18 '13 at 9:04
If you can complete the proof, then there's no tricks. If you can't complete the proof, then you need to worry about tricks. – Rahul Mar 18 '13 at 9:05
Well, I mean, all I have to do is multiply both sides by -1 and flip the inequality. I guess at that point my question shifts to "What makes flipping the inequality around when multiplying by -1 trivial?" – Trevor Alexander Mar 18 '13 at 9:07
In mathematics, axioms are assumed to be true. Flipping of the inequality when multiplied by $-1$ is one of the ordering property of real numbers. – Learner Mar 18 '13 at 9:20

HINT: use the definition of global maximum
Given $f:X \rightarrow R$, $x_0$ is a global maximum if $\forall x \in X, f(x)\leq f(x_0)$

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Mathematically, we can say $\min f(x) = - \max(-f(x))$

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Combining @Rahul's comment with @Learner's comment:

1. $f$ is maximized at $x$ if $f(x)≥f(y)$ for all $y$.
2. $−f$ is minimized at $x$ if $−f(x)≤−f(y)$ for all $y$.

From the ordering property of real numbers, we can rewrite the first equation as:

• $f(x)-f(x)-f(y) \ge f(y)-f(y)-f(x) \;\;\forall y$
• $-f(y) \ge -f(x) \Leftrightarrow −f(x)≤−f(y) \;\;\forall y$

This means that if $x$ is a maximum of $f$ in equation (1), it is a minimum of $-(f)$ in equation (2).

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