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Let $f_n(x)=\left\{\begin{matrix} 0, & x<\frac{1}{n+1} \text{ or } \frac{1}{n}<x\\ \sin^2(\frac{\pi}{x}),&\frac{1}{n+1} \leq x \leq \frac{1}{n} \end{matrix}\right.$

Show that $(f_n)$ converges pointwise to a continuous $f$ in $\mathbb{R}$. Which is this $f$? Is the convergence also uniform?

How can I find to which function $f_n$ converges,when $\frac{1}{n+1} \leq x \leq \frac{1}{n}$ ?

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Hint: Your function is $0$ at $0$, and your sequence of functions is equal to zero everywhere except on a set $1/{n+1} \leq x \leq 1/n$ which becomes arbitrarily small. So for any non-zero $z \in {\mathbb R}$, what is the value of $f_n(z)$ if $n > 1/z$? So then what is $\lim_{n\to \infty} f_n(z)$? As for uniform convergence, find the maximum value of each $f_n$ and compare this to the maximum value of the limit function $f$.

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  • $\begingroup$ Could you explain me further how I can find the function at which $f$ converges pointwise? $\endgroup$ – evinda Apr 23 '14 at 21:50
  • $\begingroup$ @evinda $f$ is the zero function, because $f_n(z) = 0$ for all $n > 1/z$. $\endgroup$ – user2566092 Apr 23 '14 at 21:54
  • $\begingroup$ But what happens, if $\frac{1}{n+1} \leq x \leq \frac{1}{n}$ ? $\endgroup$ – evinda Apr 23 '14 at 22:23

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