Prove $\sup(f^2(x))\le(\sup(f(x))^2$ Prove $\sup(f^2(x))\le(\sup(f(x))^2$
I am assuming this is true, I could find no counter examples.
$f(x)\le\sup(f(x))$
$f^2(x)\le(\sup(f(x)))^2$ when $f(x)\ge0$
$\sup(f^2(x))\le(\sup(f(x))^2$ when $f(x)\ge0$
now I can't figure out how to prove the case when $f(x)<0$. Any Ideas?
 A: Counterexample: $f(x)=\ln(x)$ in $(0,1]$ with $$\infty=\sup\left(f^2(x)\right)\not\le\left(\sup f(x)\right)^2=0$$
A: Indeed, the opposite is true. $\sup(f)^2$ is the largest value of $f$ squared. This value will either be an infinity or $f(x)^2$ will have values arbitrarily close to $\sup(f)^2$. There also may be a negative value with larger absolute value than the supremum.
A: Let $S = \{f(x)\}_x$, and let $S^2 = \{s^2\}_{s \in S}$.
Then $\sup S^2 = \max(\sup S, - \inf S)^2$.
Since $s^2 \le \sup S^2$, we have $(\sup S)^2 \le \sup S^2$ and $(\inf S)^2 \le \sup S^2$, so $\sup S^2 \ge \max(\sup S, - \inf S)^2$.
Since $|s| = \max(s, -s)$, we have $|s| \le \max(\sup S, -\inf S)$, and since
$x \mapsto x^2$ is increasing on $[0, \infty)$ we have
$s^2 = |s|^2 \le \max(\sup S, -\inf S)^2$. Taking the $\sup $ of the left hand side gives the desired result.
A: If $f$ is any negative values function, then the right had side is $(\sup f)^2 = \inf |f|$ and the left had side is $\sup f^2 = (\sup |f|)^2$. Hence the inequality is false as soon as $f$ is not constant.
