Proving/Disproving Equicontinuity of F_n(x) = sin nx

[Amer]

Prove or disprove
F_n(x) = sin nx is equicontinuous

I know the definition of equicontinuous at x_0 it says for all \epsilon >0 there exist \delta>0 such that if d ( f(x_0),f(x) ) < \epsilon then
d(x_0 , x) < \delta

trying if it is equicontinuous at x_0 = 0
Given \epsilon > 0

| f(x) | < \epsilon \Rightarrow |\sin n x | < \epsilon
delta depends on epsilon and x just how i can continue

any hints or any directions

 

[Opalg]

Prove or disprove
F_n(x) = \sin nx is equicontinuous

I know the definition of equicontinuous at x_0 it says for all \epsilon >0 there exist \delta>0 such that if d ( f(x_0),f(x) ) < \epsilon then
d(x_0 , x) < \delta

trying if it is equicontinuous at x_0 = 0
Given \epsilon > 0

| f(x) | < \epsilon \Rightarrow |\sin n x | < \epsilon
delta depends on epsilon and x just how i can continue

any hints or any directions

You have written the definition of equicontinuity the wrong way round. It should say that the family $\{f_n(x)\}$ is equicontinuous if for all \epsilon >0 there exists \delta>0 such that if d(x_0 , x) < \delta then d ( f(x_0),f(x) ) < \epsilon .

If the family $\{\sin nx\}$ is equicontinuous, then the definition of equicontinuity should hold with $\epsilon = 1/2.$ Thus there should exist $\delta>0$ such that if $|x|<\delta$ then $|\sin nx < \epsilon$ for all $n.$ Now choose $n$ so that $\frac{\pi}{2n}<\delta$ and let $x = \frac{\pi}{2n}.$ Then $|x|<\delta$ but $|\sin nx| = 1$, contradicting the definition. Thus the family cannot be equicontinuous.

 

[Evgeny.Makarov]

You have written the definition of equicontinuity the wrong way round. It should say that the family $\{f_n(x)\}$ is equicontinuous if for all \epsilon &gt;0 there exists \delta&gt;0 such that if d(x_0 , x) &lt; \delta then d ( f(x_0),f(x) ) &lt; \epsilon .

Should be "... then for all n, d ( f_n(x_0),f_n(x) ) &lt; \epsilon."

 

[Amer]

Thanks very much Opalg,

 

[Amer]

at the same way i can prove that the family F_n (x) = \cos nx is not equicontinuous

Suppose that F_n is equicontinuous at x=0 so for any \epsilon &lt; 1 there exist \delta &gt;0 such that if |x| &lt; \delta then |\cos nx - 1 | &lt; \epsilon
choose n so that \frac{\pi}{2n } &lt; \delta let x = \frac{\pi}{2n}
|\cos n\left(\frac{\pi}{2n} \right) - 1 | = |1| = 1 &gt; \epsilon
Is it Ok
Thank you again (Smile)