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.

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