# Problem 7.22

Consider $$f_n:[0,\pi]\rightarrow\mathbb{R}$$ given by $$f_n(x)=sin^{n}(x)$$, for $$n\geq1$$. Prove that $$(f_n)$$ converges pointwise to $$f:[0,\pi]\rightarrow\mathbb{R}$$ given by: $f(x) = \left\{ \begin{array}{ll} 1 & \quad x = \frac{\pi}{2} \\ 0 & \quad x \neq\frac{\pi}{2} \end{array} \right.$

Let $$f_n:[0,\pi]\rightarrow\mathbb{R}$$ be given by $$f_n(x)=sin^{n}(x)$$. For $$x=\frac{\pi}{2}$$, $$f_n(\frac{pi}{2})=sin^{n}(\frac{\pi}{2})=1^{n}=1$$ for any $$n\geq1$$. If $$x\neq\frac{\pi}{2}$$, then for any $$\epsilon>0$$, and for any $$0\leq x<\frac{\pi}{2}$$, let $$N = \frac{ln(\epsilon)}{ln(2)}$$. Then given any $$n\geq N$$, \begin{aligned} \left|sin^{n}x-0\right|&= \left|sinx\right|^{n}\\ &\leq\left|sinx\right|^{N}\\ &<2^{N}=2^{\frac{ln(\epsilon)}{ln(2)}}=\epsilon\end{aligned} So by definition, if $$x<\frac{\pi}{2}$$, $$f_n(x)=sin^{n}(x)$$ converges pointwise to zero. Which means for $$0\leq x \leq\frac{\pi}{2}$$, $$f_n(x)=sin^{n}(x)$$ converges pointwise to $f(x) = \left\{ \begin{array}{ll} 1 & \quad x = \frac{\pi}{2} \\ 0 & \quad x \neq\frac{\pi}{2} \end{array} \right.$ Now, consider the contrapositive of Theorem 7.30 which says: If $$f$$ is not continuous then $$f_n:G\rightarrow\mathbb{C}$$ is not continuous or $$(f_n)$$ does not converge uniformly.. Since $$f(x)$$ is not continuous, it is the case that $$f_n(x)=sin^{n}(x)$$ is not continuous or $$(f_n)$$ does not converge uniformly. But $$f_n(x)=sin^{n}(x)$$ is continuous on $$[0,\frac{\pi}{2}]$$ so it must be that $$(f_n)$$ does not converge uniformly.