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equicontinuous

Amer

Active member
Mar 1, 2012
275
Prove or disprove
[tex]F_n(x) = sin nx [/tex] is equicontinuous

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

trying if it is equicontinuous at [tex]x_0 = 0 [/tex]
Given [tex]\epsilon > 0 [/tex]

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

any hints or any directions
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Prove or disprove
[tex]F_n(x) = \sin nx [/tex] is equicontinuous

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

trying if it is equicontinuous at [tex]x_0 = 0 [/tex]
Given [tex]\epsilon > 0 [/tex]

[tex]| f(x) | < \epsilon \Rightarrow |\sin n x | < \epsilon [/tex]
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 [tex]\epsilon >0 [/tex] there exists [tex]\delta>0 [/tex] such that if [tex]d(x_0 , x) < \delta [/tex] then [tex]d ( f(x_0),f(x) ) < \epsilon .[/tex]

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

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
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 [tex]\epsilon >0 [/tex] there exists [tex]\delta>0 [/tex] such that if [tex]d(x_0 , x) < \delta [/tex] then [tex]d ( f(x_0),f(x) ) < \epsilon .[/tex]
Should be "... then for all n, [tex]d ( f_n(x_0),f_n(x) ) < \epsilon[/tex]."
 

Amer

Active member
Mar 1, 2012
275
Thanks very much Opalg,
 
Last edited:

Amer

Active member
Mar 1, 2012
275
at the same way i can prove that the family [tex]F_n (x) = \cos nx [/tex] is not equicontinuous

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