Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$f$ 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
up vote 2 down vote accepted

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

share|cite|improve this answer

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

share|cite|improve this answer

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.