# Show that if the norms $\| \cdot \|_1$ and $\| \cdot \|_2$ are equivalent, then $(X, \| \cdot \|_1)$ is Banach iff $(X, \| \cdot \|_2)$ is Banach

Exercise :

Show that if the norms $$\| \cdot \|_1$$ and $$\| \cdot \|_2$$ are equivalent, then the space $$(X, \| \cdot \|_1)$$ is a Banach space if and only if the space $$(X, \| \cdot \|_2)$$ is a Banach space.

Attempt :

Let $$x_n$$ be a Cauchy sequence in $$(X, \| \cdot \|_2)$$. Let $$\epsilon >0$$, then $$\exists k_0 \in \mathbb N :$$ $$\|x_k - x_l \|_2 < \frac{\epsilon}{2} \; \forall \; n,l \geq k_0$$

Every term $$x_k$$ is of the form $$x_k = (x_n^k)_{n \in \mathbb N}=(x_1^k, x_2^k, \dots)$$. Thus :

$$\|x_n^k - x_n^l\|_2 < \frac{\epsilon}{2} \; \forall \; k,l \geq k_0$$

which means that the sequence of real numbers $$(x_n^k)_{n \in \mathbb N}$$ is Cauchy $$\forall n=1,2,\dots$$. Thus, it converges to some real number $$x_n$$, which means that $$\lim_{k \to \infty} x_n^k = x_n \forall n$$.

Now, for $$l \to \infty$$, we have that :

$$\|x_n^k - x_n\|_2 < \frac{\epsilon}{2} \underset{n \to \infty}{\implies} \| x_k - x\|_2 < \epsilon$$

Thus $$\lim_{k \to \infty} x_k = x$$ which means that the sequence converges in $$(X, \| \cdot \|_2)$$.

But since the norms are equivalent, it holds that :

$$c_1\|\cdot\|_1 \leq \|\cdot \|_2 \leq c_2\|\cdot \|_1$$

Thus, it would also be : $$c_1\|x_k - x_l\|_1<\frac{\epsilon}{2} \implies \|x_k - x_l\|_1 < \frac{\epsilon '}{2}$$

But that's the definition of a Cauchy sequence in $$(X, \| \cdot \|_1)$$ and thus a sequence $$(x_n)$$ is Cauchy in $$(X, \| \cdot \|_1)$$ if and only if it's Cauchy in $$(X, \| \cdot \|_2)$$. This means that the sequence would also converge in $$(X, \| \cdot \|_1)$$ if and only if it converges in $$(X, \| \cdot \|_2)$$.

Thus $$(X, \| \cdot \|_2)$$ must be complete and thus a Banach space.

Question : Is my solution mathematically rigorous enough and correct ? I would appreciate any comments or corrections as I am a beginner at Functional Analysis.

• What you are done is ok and same argument can be done when $(X,d_1),(X,d_2)$ are metric spaces on same set $X$ and for some positive numbers $c_1,c_2$ we have $c_1d_1≤d_2≤c_2d_2$. Now if $X$ is complete w.r.t. one of metric then it is complete w.r.t. other metric also. – Sumanta Das Oct 21 '18 at 19:11
• This is what I've shown. I am asking if my approach is legit. – Rebellos Oct 21 '18 at 19:12

## 1 Answer

You seem to be assuming that the elements of $$X$$ are sequences, and there are no grounds for that assumption in the way the question is phrased. And you are trying to prove that the limit of the Cauchy sequence in $$X$$ exists, which again you don't need because it is in your hypothesis.

So, here is an argument. Suppose that $$\|x\|_1\leq a\|x\|_2\leq b\|x\|_1$$ for $$a,b>0$$. Assume that $$(X,\|\cdot\|_1)$$ is Banach. If $$\{x_k\}$$ is Cauchy in $$\|\cdot\|_2$$, then from $$\|x_k-x_j\|_1\leq a\|x_k-x_j\|_2$$ we obtain that $$\{x_k\}$$ is Cauchy in $$(X,\|\cdot\|_1)$$, which is Banach. So there exists $$x\in X$$ with $$\|x_k-x\|_1\to0$$. Now $$\|x_k-x\|_2\leq b\|x_k-x_1\|_1\to0,$$ so $$x_k\to x$$ in $$(X,\|\cdot\|_2)$$. As we can do this for any Cauchy sequence, $$(X,\|\cdot\|_2)$$ is Banach.

The converse can be obtained by reversing the roles of $$\|\cdot\|_1$$ and $$\|\cdot\|_2$$.

• Very nice and shortly stated proof, thanks a lot ! I guess I got confused, it's still the beggining so I'm trying to grasp on the whole idea. – Rebellos Oct 21 '18 at 22:47
• Don't worry. It is very easy to get confused when one is exposed to these ideas for the first time. – Martin Argerami Oct 21 '18 at 22:59