7.22
7.22 Consider \(f_n:[0,\pi]\rightarrow \mathbb{R}\) given by \(f_n(x)=sin^n(x)\), for \(n \geq 1\). Prove that \((f_n)\) converges pointwise to \(f:[0,\pi]\rightarrow\mathbb{R}\) given by: \[f(x) = \left\{ \begin{array}{lr} 1 & \text{ if } x=\frac{\pi}{2}, \\ 0 & \text{ if } x\neq\frac{\pi}{2}, \\ \end{array} \right.\] yet this convergence is not uniform.
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When \(x=\frac{\pi}{2}\) we have \(sin^n(\frac{\pi}{2})=(1)^n=1\) for all \(n\geq1\) which means that the sequence is to 1 for all values and is thus convergent.
When \(x\neq \frac{\pi}{2}\) on \([0,\pi]\), we have that \(0\leq sin^n(x)\leq1\), which by Example 7.7 on page 119, the sequence converges to 0.
We know that \(f(x)\) is not continuous since the function values are either \(0\) or \(1\). We know that \(f_n(x)=\)sin\(^n(x)\) is continuous for \(n \geq 1\). Therefore, \(f_n(x)=\)sin\(^n(x)\) does not converge uniformly by the contrapositive of Theorem 7.30.