2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.