Suppose we have the strictly monotonic function g(x) = f(x)*f(x) on all R. f(x) : R -> R. Is it possible to imagine the f(x) that reaches its infinum (inf(f)) or supremum (sup(f))?
Square root of strictly monotonic function
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real-analysis
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1Yes, if you don’t require $f$ to be continuous: let $f(0) = -1$ and $f(x) = e^x$ if $x \ne 0$. Clearly $f$ attains its minimum at $x=0$, and $g(x)=e^{2x}$ is strictly monotonic. – 2011-07-11
2 Answers
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If $f(x)=e^x$ when $x\neq 0$ and $f(0)=-1$, then $f$ reaches its infimum and $g(x)=e^{2x}$ is strictly monotone.
Similarly, if $f(x)=-e^x$ when $x\neq 0$ and $f(0)=1$, then $f$ reaches its supremum.
However, it is not possible for both to occur. This is because $|f|=\sqrt{g}$ is strictly monotone, so its supremum is not reached, and this supremum is either the supremum of $f$ or minus the infimum of $f$ (or both).
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Since $g(x)$ is monotonic and positive, so is $\sqrt{g(x)}$.
Then for each $x \in R$ we have $f(x)= \pm g(x)$. Keep in mind that the sign depends on $x$.
It is easy to see from here that :
$\inf f \geq - \sup \sqrt{g(x)} $ $\sup f \leq \sup \sqrt{g(x)} $
with equality being attained for the obvious choices of $f$.