prove $(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$

Question

prove $(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$.

$|f(x)|\geq f(x)$

$\sup{|f(x)|}\geq\sup{f(x)}$

$(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$ when $\sup{f(x)}\geq0$ ...

I don't know how to prove the other half where $\sup{f(x)}<0$. Any ideas or new approaches?

@BaronVT, the OP's proof does rely on $f(x)\gt0$ to justify maintaining the direction of the inequality when the expressions on each side are squared. — Barry Cipra, Nov 12 '14 at 19:16
My previous comment was in reply to a comment that has been removed. I would remove mine as well, but for some reason I've lost the ability to do so. (Also, I should have written "$\sup f(x)\gt0$," not just "$f(x)\gt0$.") — Barry Cipra, Nov 12 '14 at 19:28

score 1 · Answer 1 · answered Nov 12 '14 at 18:57

For any two functions $f$ and $g$, we have

$$\sup f(x)+\sup g(x)\ge\sup(f(x)+g(x))$$

We also have $|f(x)|+f(x)\ge0$ for all $x$. Therefore

$$\sup|f(x)|+\sup f(x)\ge\sup(|f(x)|+f(x))\ge0$$

Using the inequality $\sup|f(x)|\ge\sup f(x)$, which the OP proved, it follows that

$$(\sup|f(x)|)^2-(\sup f(x))^2=(\sup|f(x)|-\sup f(x))(\sup|f(x)|+\sup f(x))\ge0$$

(Remark: The OP was rightfully concerned that $a\gt b$ does not automatically imply $a^2\gt b^2$ if $b\lt0$.)

score 1 · Answer 2 · answered Nov 12 '14 at 23:58

1

If $\sup f(x)<0$, then $\inf f(x)<0$ too. Moreover $\sup|f(x)|=|\inf f(x)|$. Then…

answered Nov 12 '14 at 23:58

egreg

238,574

score 0 · Answer 3 · edited Jul 08 '20 at 22:36

Conjecture:

$(\sup|f(x)|)^2 ≥ (\sup f(x))^2$

Proof: \begin{align} |f(x)|&\geq f(x) \\ \sup(|f(x)|)&\geq \sup(f(x))\\ (\sup(|f(x)|))^2&\geq (\sup(f(x)))^2 \; \text{when}\; \sup \; (f(x))\geq 0 \end{align} Second half: \begin{align} \sup(|f(x)|)&+\sup(f(x))\geq 0 \\ \sup(|f(x)|)&\geq -\sup(f(x))\\ \sup(f(x)) < 0 &\xrightarrow{} -\sup(f(x))\geq 0\\ (\sup(|f(x)|))^2&\geq (\sup(f(x)))^2 \; \text{when}\;\sup(f(x))<0 \end{align}

prove $(\sup{|f(x)|})^2\geq(\sup{f(x)})^2$

3 Answers3