# $||.||_1$, $||.||_2$ two equivalent norms, conclude $(X, ||.||_1)$ is Banach $\iff$ $(X, ||.||_2)$ is Banach

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

## 1 Answer

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.