# Thread: Norms and Cauchy Sequences

1. ## Norms and Cauchy Sequences

Hi!

Two norms $\displaystyle ||\bullet ||$ and $\displaystyle |||\bullet |||$ on a vector space V are called equivalent if there exist 0 < a <= b such that

$\displaystyle a ||\bullet || \le |||\bullet ||| \le b ||\bullet ||$

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$\displaystyle ||\bullet ||$ and $\displaystyle |||\bullet |||$ are equivalent
Show that

A sequence is a cauchy-sequence on $\displaystyle ||\bullet ||$ <=> A sequence is a cauchy-sequence on $\displaystyle |||\bullet |||$

I do not have any clue. Any help would be much appreciated.

best regards
Rapha

2. Hello,

If $\displaystyle (a_n)_{n \in \mathbb{N}}$ is a Cauchy sequence with the $\displaystyle \|.\|$ norm, then write the definition :

$\displaystyle \forall \varepsilon' >0,~ \exists N \in \mathbb{N},~ \text{such that } \forall m,n > N,~ \|x_m-x_n\|< \varepsilon'$

But we know that there exists b such that :
$\displaystyle b \|.\| \geqslant |||.|||$
that is $\displaystyle \|.\| \geqslant \frac 1b \cdot |||.|||$ (which is possible since b>0)

So going back to the definition, we have :
$\displaystyle \frac 1b |||x_m-x_n||| \leqslant \|x_m-x_n\|< \varepsilon'$

Hence :
$\displaystyle |||x_m-x_n|||< b \varepsilon'$

Let $\displaystyle \varepsilon=b \varepsilon'$ and it'll give the Cauchy sequence definition for the $\displaystyle |||.|||$ norm.

3. Hey Moo,

thank you so much, you are awesome!