0
$\begingroup$

Not understanding the concept well, I am trying to determint the pointwise and uniform convergence of the following sequence of function:

$$f_n(x) = \frac{\sin{nx}}{n^3}, x \in \mathbb{R}$$

The only part I understand so far is that I need $\lim_{x\to\infty}{f_n(x)}$ in which I have determined that (if I am correct):

$$\lim_{x\to\infty}{f_n(x)} = \lim_{x\to\infty}{\frac{\sin{nx}}{n^3}} = 0$$

Where do i go from here?

$\endgroup$
  • $\begingroup$ It's not the function that converges but the sequence of function. Do you want to study the sequence or the series ? $\endgroup$ – Atmos Jan 21 '18 at 17:04
  • $\begingroup$ @Atmos I believe it's the sequence. $\endgroup$ – Omari Celestine Jan 21 '18 at 17:08
  • 1
    $\begingroup$ pointwise convergence is okey it also converges to zero function uniformly $\endgroup$ – daulomb Jan 21 '18 at 17:26
  • $\begingroup$ $\left|\sin\right|\leq 1$. $\endgroup$ – Jack D'Aurizio Jan 21 '18 at 17:51
1
$\begingroup$

With what you wrote I think that you want to study the sequence $\left(f_n\right)_{n \in \mathbb{N}}$. With what you wrote, you have proved that the sequence $\left(f_n\right)_{n \in \mathbb{N}}$ converge pointwisely on $\mathbb{R}$ to the function $ \ f: x \mapsto 0$. Now you need to prove the following statement for proving that it converges uniformly : $$ \left\|f_n-f\right\|_{\infty,\mathbb{R}} \underset{n \rightarrow +\infty}{\rightarrow}0 $$ Here $f=0$ and for all $x \in \mathbb{R}$ $$ \left|f_n\left(x\right)-f\left(x\right)\right|=\left|\frac{\sin\left(nx\right)}{n^3} \right|\leq \frac{1}{n^3} $$ It is true for all $x \in \mathbb{R}$ so in particular for the upper bound $$ \left\|f_n-f\right\|_{\infty, \mathbb{R}}\leq \frac{1}{n^3} $$ Hence

$$ \left\|f_n-f\right\|_{\infty,\mathbb{R}} \underset{n \rightarrow +\infty}{\rightarrow}0 $$ The convergence is hence uniform.

$\endgroup$
  • $\begingroup$ So basically I would have to find an upper limit that applies for all $x$? $\endgroup$ – Omari Celestine Jan 23 '18 at 15:17
  • 1
    $\begingroup$ Yes and that does not depend on $x$. If it tends to $0$, hence the convergence is uniform. Be careful, if $f$ was for example $-4x$ then you should have studied the upperbound of $\displaystyle x \mapsto \left|f_n(x)-4x\right|$. $\endgroup$ – Atmos Jan 23 '18 at 15:33
  • $\begingroup$ Would it be a similar solution for $f_n(x) = \frac{\sin{(nx)}}{\sqrt{n}}$? $\endgroup$ – Omari Celestine Jan 24 '18 at 16:53
  • $\begingroup$ If you are studying the sequence, yes. $\endgroup$ – Atmos Jan 24 '18 at 16:58
  • $\begingroup$ So if I am correct it would be almost the same solution just the difference being the original function? $\endgroup$ – Omari Celestine Jan 24 '18 at 17:01
1
$\begingroup$

Roughly, the deciding factor in uniform convergence is whether the "rate" of convergence of $f_n(x)$ to its limit is the same for all $x$.

A counterexample: let $f_n(x)=x/n$. The functions $f_n$ converge to $0$ pointwise. However, the time you have to wait for $f_n(x)$ to get within $0.01$ of $0$ depends on $x$. For example, when $x=3$, then the smallest $n$ such that $|f_n(x)|<0.01$ is $n=300$, while for $x=30$, the smallest such $n$ is $n=3000$. As $x$ gets further from $0$, the required $n$ to get $f_n(x)$ within $0.01$ of $0$ gets arbitrarily large, which means $f_n$ does not converge uniformly.

Back to your problem. The function $f_n(x)=\sin(nx)/n^3$ is more complicated, so given $\epsilon$, we cannot find the exact smallest $n$ for which $|f_n(x)|<\epsilon$. However, all we need to do is note that $|\sin nx|\le 1$ always, so that $|f_n(x)|\le 1/n^3$. Note that this bound does not depend on $x$; it gives us a guarantee that after $n$ is large enough, $f_n(x)$ will be close to $0$ for all $x$. Therefore, $f_n$ does indeed converge uniformly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.