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Cauchy sequence: A sequence of real numbers is a Cauchy sequence if $\forall \epsilon >0, \exists N$ s.t $\forall n,m > N, |x_n - x_m| < \epsilon$

Now, for $n \in \mathbb{N^*}$, $f_n : [0,1] \rightarrow \mathbb{R}$ is defined by $f_n(t) = t^{\dfrac{3}{n}}$

Now I am asked the following:is $\{f_n\}_{n \in \mathbb{Z}_{>0}}$ a Cauchy sequence in the vector space of continuous functions $C( [0,1 ], \mathbb{R})$ with the norm $\|x\|_p$

  • with $p = 1$

  • with $p = \infty $

My attempt:

I have found $\|f_n- f_m\|_1 = \dfrac{n}{3+n} - \dfrac{m}{3+m}$ and $\|f_n-f_m\|_{\infty} = \Big(\dfrac{n}{m}\Big)^{m/(m-n)}- \Big(\dfrac{n}{m}\Big)^{n/(m-n)}$ on a previous question.

But I have no idea how to show (or refute) that they are Cauchy sequences.

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  • $\begingroup$ something is weird with your findings at the end, why does $m$ appear in the formula? $\endgroup$ – Jorge Fernández Hidalgo Jan 22 '17 at 18:02
  • $\begingroup$ It would help you to find a formula for $||f_n-f_m||$ $\endgroup$ – Jorge Fernández Hidalgo Jan 22 '17 at 18:02
  • $\begingroup$ I'm sorry, those were $ ||f_n-f_m||_1$ and $||f_n-f_m||_{\infty}$ $\endgroup$ – Sylvester Stallone Jan 22 '17 at 18:11
  • $\begingroup$ oh, the first formula is easy to work with. $\endgroup$ – Jorge Fernández Hidalgo Jan 22 '17 at 18:12
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The first one is not a cauchy sequence, because $f_n$ converges pointwise to a discontinuous sequence, so it cannot converge uniformly by the uniform limit theorem.

For the second part you can note that $||f_n-f_m||_1=\frac{1}{1+\frac{3}{n}}-\frac{1}{1+\frac{3}{m}}=\frac{n}{n+3}-\frac{m}{m+3}$ when $n>m$. This is clearly less than $1-\frac{m}{m+3}=\frac{3}{m+3}$.

So it is sufficient to take $N$ such that $\frac{3}{N+3}<\epsilon$ to prove the sequence is Cauchy.

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