Question on infimum and supremum

Question

How to find infimum and supremum of a complex function like $\frac{x}{\sin(x)}$ in the interval $(0,\pi/2]$.

Answer 1

First of all, infimum and supremum can be defined only on partially ordered sets. So there is no infimum or supremum for complex valued function, unless you define partial ordering on complex numbers.

However, function $f(x)=x/\sin x$ is real on $I=(0,\pi/2]$, so we can interpret it as a real valued function.

Function $f(x)$ is continuous on $I$. So all we need is to explore whether the function has global minimum/maximum inside the interval and take limits at the ends of the interval. Can you do it?

Answer 2

The derivative $(\sin x-x \cos x)/(\sin ^2 x)$ can only be zero if $x=\tan x$ which doesn't happen in $(0,\pi/2].$ [This is fairly easy to show, one way is to note the derivative of $\sin x$ is at most $1$ etc.] Plugging in one value to this derivative then shows your function is strictly increasing on $(0,\pi/2].$ Note also that as $x \to 0^+$ the function goes to $1.$

So your function has no minimum (but its infimum is 1 as $x \to 0^+$) and its max occurs when $x=\pi/2.$ In between there are no local max/min.