Exercise 7.5

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Exercise 5: 
    Let $$f_n(x)=
    \begin{cases}
        0 & \displaystyle \biggl(x<\frac{1}{n+1}\biggr), \
        \displaystyle \sin^2\frac{\pi}{x} & \displaystyle \biggl(\frac{1}{n+1}\le x\le\frac{1}{n}\biggr), \
        0 & \displaystyle \biggl(\frac{1}{n}<x\biggr).
    \end{cases}$$
    Show that $\{f_n\}$ converges to a continuous function, but not uniformly. Use the series $\sum f_n$ to show that absolute convergence, even for all $x$, does not imply uniform convergence.

analambanomenos · Accepted Answer

\def\eps{\varepsilon} For $x\le 0$, we have $f_n(x)=0$ for all $n$, and for $x>0$ and $n$ large enough so that $n^{-1}N$, so $\{f_n(x)\}$ does not converge uniformly to $f(x)$. Let $$F_N(x)=\sum_{n=1}^N f_n(x)= \begin{cases} 0 & \displaystyle \biggl(x<\frac{1}{N+1}\biggr), \ \displaystyle \sin^2\frac{\pi}{x} & \displaystyle\biggl(\frac{1}{n+1}\le x\biggr). \end{cases}$$ Then $F_N(x)$ converges to $$F(x)= \begin{cases} 0 & \displaystyle (x\le0), \ \displaystyle \sin^2\frac{\pi}{x} & \displaystyle(0

Exercise 7.5

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Exercise 7.5

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