Thread: Cauchy Sequences

Problems:
For each positive integer k define

a.$\displaystyle f_{k}(x)=x^k$
b.$\displaystyle f_{k})(x)=e^{x/k}$
c. $\displaystyle f_{k}(x)=cos(x/k)$

for $\displaystyle 0 \leq x \leq 1$. Is the sequence $\displaystyle { f_{k}:[0,1] \rightarrow R}$ a Cauchy Sequence in the metric space $\displaystyle C([0,1]),R$

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By definition, a sequence $\displaystyle {f_{k}} $ is Cauchy sequence in $\displaystyle C([0,1]),R$ if and only if there is an $\displaystyle \epsilon > 0 $

such that

$\displaystyle |f_{k}(x)-f_{l}(x) | < \epsilon$ for all x in [a, b]

For a I said that $\displaystyle f_{k}(x)=x^k$ is only Cauchy for when $\displaystyle k < 1 $ since the distance between the $\displaystyle f_{k}'s $ get smaller and smaller together. Thus this sequence is a Cauchy sequence.

For b, $\displaystyle f_{k})(x)=e^{x/k}$, I said that this is not Cauchy as it is distance is getting larger and larger since the sequence of function grows exponentially. So as k and l goes to infinity, $\displaystyle |f_{k}(x)-f_{l}(x) | $ does not go to zero.

For c, $\displaystyle f_{k}(x)=cos(x/k)$, I said that this is Cauchy with a similar argument with part (a).

I have no idea how to put this in a formal proof though, but the problem seems like it only wants a true or false type of answer, I think.

Thank you for your time.

Originally Posted by Paperwings

a.$\displaystyle f_{k}(x)=x^k$

If this was a uniform Cauchy sequence on $\displaystyle [0,1]$ then it would be a uniformly convergent sequence on $\displaystyle [0,1]$. If it was a uniformly convergent sequence on $\displaystyle [0,1]$ then the limit function would be continous. But it is not.