I want to show if the following sequence of functions converges pointwise and converges uniformly on R? $ f_n(x) = \sin(x/n) $

Here's what I did: -Fix $\displaystyle x\in R, \lim_{n\rightarrow\infty}f_n(x) = \lim_{n\rightarrow \infty} \sin(n/x) = 0 $. so $f_n $ converges to f s.t. f(x) = 0 for all x in R, point wisely. Now for uniform convergence: $\|f_n - f\|_{\sup} = \sup_R|f_n - f|= \sup_R|\sin(x/n)-0| = \sup_R|\sin(x/n)|\rightarrow0 $ as n$\rightarrow\infty$, so it's uniformly convergent. Is this correct? and should I right anything more for the uniform convergence?


1 Answer 1


No. Please observe, that for every $n$ there exists $x$ such that $\sin(x/n)=1$.

Probably you want to prove that function converge uniformly on every fixed, bounded interval. It is true, but uniform convergence on the whole real line — not.

  • $\begingroup$ So, $f_n(x)$ is not convergent pointwisely on R? $\endgroup$
    – Jack
    Commented Feb 28, 2018 at 18:33
  • 1
    $\begingroup$ @Jack- It is convergent pointwise to $0$. However, it is not uniformly convergent to $0$. Those are two separate notions $\endgroup$
    – user67803
    Commented Feb 28, 2018 at 19:02

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