I have this question I have been cracking for hours I need help. Preface: not a homework question, just trying to gain some intuition on this equicontinuous property.

Given $f_n(x) = (\frac{x}{x+1})^n \sin(x), x > 0$, is it

1) uniformly convergent,

2) pointwise equicontinuous,

3) uniformly equicontinuous

over $(0,\infty)$? What about $(0,1)$?

1) $f_n(x) = (\frac{x}{x+1})^n \sin(x)= (1-\frac{1}{x+1})^n \sin(x)$

The pointwise limit of this question is $f(x) = 0$, since $\left|(1-\frac{1}{x+1})\right|<1, \forall x > 0$

Try to show $\forall \epsilon > 0, \exists N > 0$ such that $|f_n(x) - f(x)| < \epsilon, \forall n > N$

Then $|f_n(x) - f(x)| = |f_n(x) - 0| = |f_n(x)| = \left|\left(\frac{x}{x+1}\right)^n \sin(x)\right| \leq \left|\frac{x}{x+1}\right|^n$

^ Here how do I wrap this up?

2) A sequence $f_n$ is pointwise equicontinuous if $\forall x \in (0,\infty)$, $\forall \epsilon > 0, \exists \delta > 0$ s.t. $\forall y \in (0,\infty)$ $\forall n \in \mathbb{N}, |x-y| < \delta \implies |f_n(x) - f_n(y)| < \epsilon$

Let $\epsilon >0$ be given, then $|f_n(x) - f_n(y)| = \left|\left(\frac{x}{x+1}\right)^n \sin(x) - \left(\frac{y}{y+1}\right)^n \sin(y)\right| \leq \left|\left(\frac{x}{x+1}\right)^n- \left(\frac{y}{y+1}\right)^n\right|$

^ Stuck, how to proceed?

3) A sequence $f_n$ is uniformly equicontinuous if $\forall \epsilon > 0, \exists \delta > 0$ s.t. $\forall n \in \mathbb{N}$, $\forall x,y \in (0,\infty)$, $|x-y| < \delta \implies |f_n(x) - f_n(y)| < \epsilon$

Not even sure how to start for this one. What is the difference between pointwise and uniform equicontinuous?

Thanks a bunch!!

(I hope the problem wasn't too difficult)


Right, in 1) we have pointwise convergence to $0.$ But note that $f_n\to 0$ uniformly on $(0,\infty)$ iff

$$\tag 1 \sup_{(0,\infty)}|f_n| \to 0.$$

But for each $n$ the $\sup$ in $(1)$ is $1.$ (To see this, fix $n$ and consider $f_n(\pi/2+ 2\pi m)$ as $m\to \infty.$)

3) $(f_n)$ is uniformly equicontinuous on $(0,\infty).$ Proof: Let $b_n(x) = (1-1/(x+1))^n.$ Then

$$|b_n(x)\sin x - b_n(y)\sin y| \le |b_n(x)\sin x - b_n(x)\sin y| + |b_n(x)\sin y - b_n(y)\sin y|$$ $$ \le |\sin x - \sin y| + |b_n(x) - b_n(y)|.$$

Now $\sin x$ is uniformly continuous on $\mathbb R.$ Thus it suffices to show $b_n(x)$ is uniformly equicontinuous on $(0,\infty).$ To do this, it suffices to show that $b_n'$ is a uniformly bounded sequence on $(0,\infty).$ (This will show that $(b_n)$ is uniformly Lipshitz, which gives uniform equicontinuity in spades.)

To show $b_n'$ is uniformly bounded, I found the unique solution of $b_n''(x) = 0$ is $x=(n-1)/2.$ We are then left contemplating

$$b_n'((n-1)/2) = n[1-1/((n-1)/2 + 1)]^{n-1}[(n-1)/2+1]^{-2}.$$

Verify that this sequence of numbers $\to 0$ and we're done. (So not only is $b_n'$ uniformly bounded, $b_n'\to 0$ uniformly on $(0,\infty).$)

  • $\begingroup$ Can you comment on the parts about equicontinuity? Or is it that failure of uniform convergence implies non equicontinuity? $\endgroup$
    – Olórin
    Feb 18 '16 at 1:45
  • $\begingroup$ No, we can have equicontinuity even when uniform convergence fails. By the way you made a mistake in the last inequality of 2). I believe uniform equicontinuity holds on $[0,\infty)$, I"ll try to write something later. $\endgroup$
    – zhw.
    Feb 19 '16 at 17:57

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