# 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, 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.

However, it seems to me that math is all about a meticulous examination of surprising subtleties. Does anyone know of a good way to prove this statement?

• $f$ 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
• 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?" – bright-star 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 $\forall x \in X, f(x)\leq f(x_0)$

$f$ is maximized at $x$ if $f(x)\geq f(y)$, for all $y$. Similarly, $−f$ is minimized at $x$ if $−f(x) \leq −f(y)$, for all $y$.

Let's now add to the left and right sides of the inequality $−f(x) \leq −f(y)$ respectively $f(y)-f(y) = 0$ and $f(x)-f(x) = 0$, that is

$$(f(y)-f(y))-f(x) \leq (f(x)-f(x))-f(y)$$

which can be written as

$$f(y) + (-f(y)-f(x)) \leq f(x) + (-f(x)-f(y))$$

we can remove $(-f(y)-f(x))$ from both sides to obtain

$$f(y) \leq f(x)$$

This proof followed from the ordering property of real numbers.

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