EDITTED: Find all values of $a$ and $b$ so that $ax^n+b\cos\left(\frac{x}{n}\right)$ is Cauchy.

Question

For each $n\in\mathbb{N}$ let

$$f_n(x)=ax^n+b\cos\left(\frac{x}{n}\right), \text{ } x\in[0,1].$$

Find all values of $a$ and $b$ for which $(f_n)$ is a Cauchy sequence in $C[0,1]$, the space of continuous functions on $[0,1]$ equipped with the metric

$$d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|.$$

EDITTED:

My working:

For any given $\epsilon>0$ we need to find an integer $N$ such that for all $m,n\geq N\implies d(f_m,g_n)<\epsilon$.

So we have
$d(f_m,g_n)=\sup_{x\in[0,1]}|f_m(x)-g_n(x)|$
$=\sup_{x\in[0,1]}|a(x^n-x^m)+b(\cos\left(\frac{x}{m}\right)-\cos\left(\frac{x}{n}\right))|$

I get stuck if I used the definition of Cauchy.

As mentioned by Martín-Blas Pérez Pinilla, the space $C[0,1]$ is complete, hence Cauchy implies convergent and conversely?

My doubt is, since the metric is the supremum metric, is the convergence that we are talking about the uniform convergence?

But we don't know the limit apriori yet, then how can we show that it is convergent?

Can anybody please give some helps on how to proceed?

Thanks in advance! Really appreciate it!

Answer 1

Hint: your space is complete, so Cauchy $\implies$ convergent. Now, what happens with the pointwise limit?