# equicontinuous

#### Amer

##### Active member
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

##### MHB Oldtimer
Staff member
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

##### Well-known member
MHB Math Scholar
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 $$d ( f(x_0),f(x) ) < \epsilon .$$
Should be "... then for all n, $$d ( f_n(x_0),f_n(x) ) < \epsilon$$."

#### Amer

##### Active member
Thanks very much Opalg,

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#### Amer

##### Active member
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 $$\epsilon < 1$$ there exist $$\delta >0$$ such that if $$|x| < \delta$$ then $$|\cos nx - 1 | < \epsilon$$
choose n so that $$\frac{\pi}{2n } < \delta$$ let $$x = \frac{\pi}{2n}$$
$$|\cos n\left(\frac{\pi}{2n} \right) - 1 | = |1| = 1 > \epsilon$$
Is it Ok
Thank you again