I know how to prove the "$\leq$", but not the "$\geq$"
2026-03-30 12:24:18.1774873458
Prove that, if $f:X\to[0,\infty)$ then $\sup([f(x)]^2)=[\sup(f(x))]^2$
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Take sequence $x_n$ s.t. $f(x_n) \to \sup f(X)$. Consider $f(x_n)^2$, then $f(x_n)^2 \to (\sup f(X))^2$ by continuity of squaring (treat the case $\sup f(X) = \infty$ separately). Thus $\sup \{f(x)^2 \mid x \in X\} \geq (\sup f(X))^2$.