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$(f_n)$$_n$$_\in $$_\mathbb N$ is a sequence of functions where $f_n : [0,2\pi] \to \mathbb R$ $\ \forall n \in \mathbb N$. Find all values of $x \in [0,2\pi]$ such that $(f_n)$$_n$$_\in $$_\mathbb N$ converges and find pointwise limit if it exists.

(i) $f_n (x) = $sin$ (\frac{x}{n})$

(ii)$f_n (x) = $sin$ (nx)$

(iii)$f_n (x) = $sin$^n$ $(x)$

(i) This is easy to show: $lim_{n \to \infty} f_n(x) = f(x)$ where $f(x) = 0 \ \ \forall x \in [0,2\pi]$

So $f_n(x)$ converges $\forall x \in [0,2\pi]$

(ii) $f_n(x)$ converges if $x = 0, \pi, 2\pi$. And the pointwise limit is $f(x) = 0$ if $x = 0, \pi, 2\pi$

(iii) $$f_n(x) = \begin{cases} 0, & \text{if $x = 0, \pi, 2\pi$} \\[2ex] sin^n (x), & \text{if $x\neq 0, \pi, 2\pi$} \end{cases}$$

Then we have

$$\lim_{n \to \infty} f_n(x) = f(x) = \begin{cases} 0, & \text{if $x \neq \frac{3\pi}{2} $} \\[2ex] 1, & \text{if $x = \frac{\pi}{2}$} \\[2ex] Doesn't \ exist, & \text{if $x = \frac{3\pi}{2}$} \end{cases}$$

So $f_n(x)$ converges if $x \in [0, 2\pi]$ \ {$\frac{3\pi}{2}$}

$$$$

I somewhat feel that my answers for (ii) and (iii) are wrong.

Any comment / correction is appreciated

Community
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TUC
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1 Answers1

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I do not know if you need to prove your answers. If that is the case, there are many steps missing. In any case, (ii) is correct. (iii) is basically an exponential, for that to converge the base belongs to $(-1,1]$, in your case $-1<\sin(x)\leq1$ which implies $x\neq3\pi/2$, and convergence is as you said.

Eduardo Elael
  • 341
  • No justification / proof is needed. I just wanted to practice more on the concept of sequence of function. – TUC Nov 21 '18 at 15:10
  • for point-wise convergence you are basically looking into different sequences (one for each x). For a better concept of convergence of sequence of functions, take a look into Hilbert Spaces. – Eduardo Elael Nov 21 '18 at 15:15
  • So if I understand this correctly, for (iii), if $\pi \le x \le 2\pi$, then it oscillates between positive and negative values with $n \in \mathbb N$ so we only consider $0 \le x \le \pi$. But for pointwise limit, is it 0 $\forall x \in [0, 2\pi]$ \ {$\frac{\pi}{2}$} or do we only consider $x \in [0,\pi]$ for pointwise limit as well? – TUC Nov 21 '18 at 15:18
  • Actually, your answers are correct, I overlooked the question, it only oscillates without decaying for sin(x)= -1. Otherwise, although it oscillates it converges to zero. I edited the answer to corroborate with yours. – Eduardo Elael Nov 21 '18 at 15:46