Is this a Cauchy sequence?

Question

Consider a sequence of function defined on a compact interval $[-k,k]$ of $\mathbb{R}$ by $$f_n(x) := \cos\left(\frac{x}{2n}\right)\text{sinc}\left(\frac{x}{2n}\right),$$ where $\text{sinc}(y) := \frac{\sin(y)}{y}$.

Given $\varepsilon > 0$, is it possible to find $N \in \mathbb{N}$ such that $$ m, n \geq N ~~\Rightarrow~~ \big|f_n(x) - f_m(x)\big| < \varepsilon, ~~~~~~\forall x \in [-k,k].$$

My idea : $$\sup\limits_{x \in [-k,k]}\big|f_n(x) - f_m(x)\big| = \sup\limits_{x \in [-k,k]}\left|\cos\left(\frac{x}{2n}\right)\text{sinc}\left(\frac{x}{2n}\right) - \cos\left(\frac{x}{2m}\right)\text{sinc}\left(\frac{x}{2m}\right)\right| \\= \sup\limits_{x \in [-k,k]}\left|\cos\left(\frac{x}{2n}\right)\frac{\sin\left(\frac{x}{2n}\right)}{\frac{x}{2n}} - \cos\left(\frac{x}{2m}\right)\frac{\sin\left(\frac{x}{2m}\right)}{\frac{x}{2m}}\right| \\= \sup\limits_{x \in [-k,k]}\left|\frac{2\sin\left(\frac{x}{2n}\right)\cos\left(\frac{x}{2n}\right)}{\frac{x}{n}} - \frac{2\sin\left(\frac{x}{2m}\right)\cos\left(\frac{x}{2m}\right)}{\frac{x}{m}}\right|\\= \sup\limits_{x \in [-k,k]}\left| \frac{\sin\left(\frac{x}{n}\right)}{\frac{x}{n}}- \frac{\sin\left(\frac{x}{m}\right)}{\frac{x}{m}}\right|.$$ But then I don't know what to do...

Are you trying to prove, or check, whether for some fixed $;x\in[-k,k];$, the sequence $;{f_n(x)};$ is a Cauchy sequence? — DonAntonio, Sep 05 '16 at 20:03
Hint: Have you considered the reverse triangle inequality in the last step? — Mathemagician1234, Sep 05 '16 at 20:05
Note that $$\cos \frac{k}{2n}\operatorname{sinc} \frac{k}{2n} \leqslant f_n(x) \leqslant 1$$ for $x\in [-k,k]$ when $n \geqslant k/\pi$. — Daniel Fischer, Sep 05 '16 at 20:23
Sorry for the late reply. I'm trying to check whether ${f_n}_{n \in \mathbb{N}}$ is a Cauchy sequence. The $x \in [-k,k]$ should not be fixed. — user366489, Sep 05 '16 at 20:35

score 0 · Answer 1 · answered Sep 05 '16 at 20:08

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Assuming my question in the comments above has a positive answer:

$$\begin{align*}&x\neq m\pi\;,\;\;m\in\Bbb Z\implies f_n(x)=\cos\frac x{2n}\cdot\frac{\sin\frac x{2n}}{\frac x{2n}}\xrightarrow[n\to\infty]{}1\cdot1=1\\{}\\ &x= m\pi\;,\;\;0\neq m\in\Bbb Z\implies f_n(x)=\cos\frac x{2n}\cdot\frac{\sin\frac x{2n}}{\frac x{2n}}\xrightarrow[n\to\infty]{}1\cdot0=0\\{}\\&x=0\;\implies \text{ The function's undefined on this point}\end{align*}$$

So where the function $\;f_n(x)\;$ , for a fixed $\;x\;$ , are defined, they form a converging sequence and thus they are a Cauchy sequence.

answered Sep 05 '16 at 20:08

DonAntonio

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This shows that each $(f_n(x))_n$ is Cauchy (in $\mathbb R$). The question asks to show that $(f_n)$ is Cauchy (presumably for the uniform norm). (Yes this was not so clear at the start.) – Did Sep 05 '16 at 20:44
I take the blame for that. My question was ill-posed at first. – user366489 Sep 05 '16 at 21:04

score 0 · Answer 2 · answered Sep 07 '16 at 16:23

Both $\cos(x)$ and $\text{sinc}(x)$ are even continuous functions bounded by 1 and mapping 0 to 1. Therefore, given $0 < \varepsilon < 1$, you can always find $N \in \mathbb{N}$ big enough to ensure that $$\cos\left( \frac{x}{2N}\right)\text{sinc}^3\left(\frac{x}{2N}\right) ~<~ \frac{\varepsilon}{2}$$ for all $x \in [-k,k]$.

Now, given $m, n \geq N$, you have $$ \sup\limits_{|x| \leq k}\left| \cos\left( \frac{x}{2N}\right)\text{sinc}^3\left(\frac{x}{2N}\right) - \cos\left( \frac{x}{2N}\right)\text{sinc}^3\left(\frac{x}{2N}\right)\right| ~\leq~ \varepsilon.$$

Is this a Cauchy sequence?

2 Answers2