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I came across an equality, which states that

If $D\subset\mathbb{R}^n, n\geq 2$ is compact, for each $ f\in C(D)$, we have the following equality $$\max\limits_D |f(x)|=\max\limits\{|\max\limits_D f(x)|, |\min\limits_D f(x)|\}.$$

Actually I can not judge if it is right. Can anyone tell me if it is right, and if so, how to prove it?
Thanks a lot.

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I imagine you want the max to exist. Assuming $f$ is continuous on $D$ will do it. – André Nicolas Jun 26 '12 at 3:25
I am sorry for my mistake. I have re-edited it! – nuage Jun 26 '12 at 3:30
The result is clear. It basically says nothing. – André Nicolas Jun 26 '12 at 3:36
up vote 1 down vote accepted

I assume that you are assuming that $\max$ and $\min$ exists or that you are assuming that $f(x)$ is continuous which in-turn guarantees that $\max$ and $\min$ exists, since $D$ is compact.

First note that if we have $a \leq y \leq b$, then $\vert y \vert \leq \max\{\vert a \vert, \vert b \vert\}$, where $a,b,y \in \mathbb{R}$. Hence, $$\min_D f(x) \leq f(x) \leq \max_D f(x) \implies \vert f(x) \vert \leq \max\{\vert \max_D f(x) \vert, \vert \min_D f(x) \vert\}, \, \forall x$$ Hence, we have that $$\max_D \vert f(x) \vert \leq \max\{\vert \max_D f(x) \vert, \vert \min_D f(x) \vert\}$$ Now we need to prove the inequality the other way around. Note that we have $$\vert f(x) \vert \leq \max_D \vert f(x) \vert$$ This implies $$-\max_D \vert f(x) \vert \leq f(x) \leq \max_D\vert f(x) \vert$$ This implies $$-\max_D \vert f(x) \vert \leq \max_D f(x) \leq \max_D\vert f(x) \vert$$ $$-\max_D \vert f(x) \vert \leq \min_D f(x) \leq \max_D\vert f(x) \vert$$ Hence, we have that $$\vert \max_D f(x) \rvert \leq \max_D\vert f(x) \vert$$ $$\vert \min_D f(x) \rvert \leq \max_D\vert f(x) \vert$$ The above two can be put together as $$\max \{\vert \max_D f(x) \rvert, \vert \min_D f(x) \rvert \} \leq \max_D\vert f(x) \vert$$ Hence, we can now conclude that $$\max \{\vert \max_D f(x) \rvert, \vert \min_D f(x) \rvert \} = \max_D\vert f(x) \vert$$

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Thank you, Marivis. I have been helped! – nuage Oct 28 '12 at 11:41

It is correct. Proving it is straight forward enough, so I'll just give you a hint. $|f(x)|=f(x)$, provided that $f(x)\geq 0$. And, $|f(x)|=-f(x)$ otherwise. Now, what happens if the RHS is not equal to the first possibility? What happens if it's not equal to the 2nd possibility? Look at if the max and mins are +ve or -ve. ect, do case work on this. You'll get it.

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Since $f$ is continuous on a compact set $D$, the minimum and maximum of $f(x)$ over $D$ exist. Suppose that the minimum is $a$ and the maximum is $b$.

There are two cases, depending on the relative size of $|a|$ and $|b|$.

Case 1: $|a| \le |b|$. Then $\max |f(x)|=|b|$. (Actually, the absolute value sign around $b$ is unnecessary, since if $|a|\le |b|$ then $b$ must be $\ge 0$.)

But $|\max f(x)|=|b|$ and $|\min f(x)|=|a|$, so $\max\{|\max f(x)|, |\min f(x)|\}=|b|$.

Case 2: $|a| \gt |b|$. This can only happen if $a$ is negative. Then $\max|f(x)|=|a|$. But $|\max f(x)|=|b|$ and $|\min f(x)|=|a|$, so $\max\{|\max f(x)|, |\min f(x)|\}=|a|$.

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