Question

The function f(x)= sin^2x is uniformly continuous in the interval [0,π]. True or false

Answer 1

$\displaystyle Let\ \epsilon>0\text{ be given, we need to show that }\exists\ a\ \delta(\epsilon)>0\ \ni|\sin^2x-\sin^2y|<\epsilon\ \text{whenever}\\ |x-y|<\delta(\epsilon)\ and\ x,y\in[0,\pi].\\ \text{Now, let }\epsilon>0\text{ be given. Then, whenever}\ |x-y|<\delta(\epsilon)\ and\ x,y\in[0,\pi]\ \text{we have}:\\ |\sin^2x-\sin^2y|=|(\sin x+\sin y)(\sin x-\sin y)|, \text{difference of two squares}\\ \qquad\qquad\qquad\quad=|\sin x+\sin y||\sin x-\sin y|\\ \qquad\qquad\qquad\quad\leq(|\sin x|+|\sin y|)|\sin x-\sin y|, \text{triangle inequality}\\ \qquad\qquad\qquad\quad=2|\sin x-\sin y|, \text{since }|\sin x|\leq1\ \forall \ x\in\R\\ \qquad\qquad\qquad\quad=2\times|2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})|\\ \qquad\qquad\qquad\quad=4|\cos(\frac{x+y}{2})||\sin(\frac{x-y}{2})|\\ \qquad\qquad\qquad\quad=4|\sin(\frac{x-y}{2})|\leq2|x-y|=2\times \delta=\epsilon, \text{if }\delta=\frac{\epsilon}{2}.\\ \quad\text{Thus, given any }\epsilon>0\ \exist\ \delta(\epsilon)>0\ni|\sin^2x-\sin^2y|<\epsilon\ \text{whenever}\ |x-y|<\delta(\epsilon)\ and\\\ x,y\in[0,\pi], \text{showing that }f(x)=\sin^2x \text{ is uniformly continuous on }[0, \pi].$