3
$\begingroup$

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)

$\endgroup$
0
1
$\begingroup$

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 $(0,\infty).$)

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.