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

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Discussion Overview

The discussion revolves around the equicontinuity of the family of functions defined by F_n(x) = sin(nx) and F_n(x) = cos(nx). Participants explore the definition of equicontinuity and attempt to prove or disprove whether these families of functions are equicontinuous at specific points, particularly at x_0 = 0.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to prove equicontinuity of F_n(x) = sin(nx) at x_0 = 0 using the definition of equicontinuity, noting that delta depends on epsilon and x.
  • Another participant corrects the definition of equicontinuity provided by the first participant, stating that it should indicate that for all ε > 0, there exists δ > 0 such that if d(x_0, x) < δ, then d(f(x_0), f(x)) < ε.
  • This second participant argues that if the family {sin(nx)} is equicontinuous, then for ε = 1/2, there should exist δ such that if |x| < δ, then |sin(nx)| < ε for all n. They provide a counterexample to show that this leads to a contradiction, suggesting that the family cannot be equicontinuous.
  • A third participant reiterates the correction regarding the definition of equicontinuity, emphasizing the need for the condition to hold for all n.
  • Another participant proposes a similar argument for the family F_n(x) = cos(nx), suggesting that it is also not equicontinuous by following a similar reasoning process and providing a specific counterexample.

Areas of Agreement / Disagreement

Participants generally disagree on the equicontinuity of the families of functions. While some argue that neither family is equicontinuous, the discussion remains unresolved regarding the implications of the definitions and the validity of the proofs presented.

Contextual Notes

Participants express uncertainty regarding the application of the definition of equicontinuity and the implications of their counterexamples. The discussion highlights the need for careful consideration of the conditions under which equicontinuity is evaluated.

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

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

trying if it is equicontinuous at x_0 = 0
Given \epsilon &gt; 0

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

any hints or any directions
 
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Amer said:
Prove or disprove
F_n(x) = \sin nx is equicontinuous

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

trying if it is equicontinuous at x_0 = 0
Given \epsilon &gt; 0

| f(x) | &lt; \epsilon \Rightarrow |\sin n x | &lt; \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 &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 .

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.
 
Opalg said:
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."
 
Thanks very much Opalg,
 
Last edited:
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)
 

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