Let $||.||_1$, $||.||_2$ be two equivalent norms in a vector space $X$.

a) Show that the Cauchy sequences in $(X, ||.||_1)$ and in $(X, ||.||_2)$ are the same

b) Show that $\lim_{n\to\infty}||x_n-x||_1=0\iff \lim_{n\to\infty}||x_n-x||_2=0$

c) Conclude that $(X, ||.||_1)$ is Banach $\iff$ $(X, ||.||_2)$ is Banach

I didn't understand question a). Is it asking for me to show that every cauchy sequence that converges in the first space is also a Cauchy sequence that converges in the second space?

b) $$\lim_{n\to\infty}||x_n-x||_1=0\iff \forall n_0,\exists n>n_0\implies ||x_n-x||_1<\epsilon$$

But $||.||_1$ and $||.||_2$ being equivalent means that there exists constants $a,b$ such that

$$a||.||_2\le ||.||_1 \le b||.||_2$$

so our condition above means

$$ \forall n_0,\exists n>n_0\implies a||x_n-x||_2 \le ||x_n-x||_1<\epsilon_2$$

and if we take $\epsilon_2 = 2\epsilon$ we have

$$ \forall n_0,\exists n>n_0\implies ||x_n-x||_2 \le \epsilon$$

which implies $\lim_{n\to\infty}||x_n-x||_2=0$

The converse is similar

c) If $(X, ||.||_1)$ is Banach, it means it is a vector normed space that is complete. Which means that every Cauchy sequence in $X$ converges in $X$ using the norm $||.||_1$. By b) we know that a sequence that converges to $x$ with norm $1$ also converges to $x$ with norm $2$. So now I should use a) somehow to say the sequences are the same


You are being asked to show that if $\{x_n\}$ is Cauchy with respect to one of the norms then it is Cauchy with respect to the other.

For instance suppose $\{x_n\}$ is Cauchy with respect to $\|\cdot\|_1$.

Let $\epsilon > 0$ be given.

There exists $N \in \mathbb N$ with the property that $n,m \ge N \implies \|x_n - x_m\|_1 < a \epsilon.$

Thus $n,m \ge N \implies \|x_n - x_m\|_2 < \epsilon$ so that $\{x_n\}$ is Cauchy with respect to $\|\cdot\|_2$.

This is unrelated to convergence since the space could fail at this point to be complete.


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