# Sequences are Cauchy depending on the norm?

Assume that we have two Banach spaces $B_1, B_2$ with their underlying sets being identical. Is it possible that a Cauchy sequence in one of the spaces would fail to be Cauchy in the other? I am assuming that it is possible since the Cauchy test directly depends on the norm. However, this seems kind of unnatural, will someone please clarify.

Let $V$ be a vector space of Hamel dimension $c$. Let $T$ be a linear bijection from $V$ to $\ell_1$ and let $S$ be a linear bijection from $V$ to $\ell_2$. This can be done since any two separable infinite dimensional Banach spaces have Hamel dimension $c$.

Define two norms on $V$ via $\Vert x\Vert_T=\Vert Tx\Vert_{\ell_1}$ and $\Vert x\Vert_S=\Vert S x\Vert_{\ell_2}$. Then $(V,\Vert\cdot\Vert_S)$ is isomorphic to $\ell_2$ and $(V,\Vert\cdot\Vert_T)$ is isomorphic to $\ell_1$. In particular, these two norms are complete.

Since $\ell_1$ is not isomorphic to $\ell_2$, it follows that these two norms are not equivalent. Per Julien's answer, these two norms have different Cauchy sequences.

• Unlike me, you answered the question...+1. – Julien Jun 21 '13 at 21:17
• This is a minor obstacle at most, but why is it that any two separable Banach spaces have dimension $\mathfrak c$ (and not any less)? – tomasz Jun 21 '13 at 21:19
• @tomasz I forgot an important detail (infinite dimensional, of course; corrected now)... – David Mitra Jun 21 '13 at 21:20
• @julien I wonder if an example is known without using Choice. – David Mitra Jun 21 '13 at 21:22
• @tomasz: It's a theorem of Mackey. See this post – Martin Jun 21 '13 at 21:23

Edit: here is a slightly different approach for a counterexample, without reference to any exterior Banach space. We claim that for every infinite-dimensional Banach space $(X,\|\cdot\|)$, there exists another complete norm $\|\cdot \|'$ on $X$ which is not equivalent to the original norm and therefore has different Cauchy sequences.

Let $\{x_j\,;\, j\in J\}$ be a Hamel basis for $X$. Without loss of generality, we can assume that $\|x_j\|=1$ for every $j\in J$. We know by Baire that $J$ is not countable. But all we need is that there exists $\mathbb{N}^*\simeq J_0\subseteq J$. Now consider the linear operator $T:X\longrightarrow X$ defined on the basis by $Tx_n:=nx_n$ for all $n\in\mathbb{N}^*\simeq J_0$ and $T_jx_j:=x_j$ for $j\in J\setminus J_0$. By construction, $T$ is invertible and unbounded.

Define the new norm by $\|x\|':=\|Tx\|$. Then $T:(X,\|\cdot\|')\longrightarrow (X,\|\cdot\|)$ is an isometric isomorphism whence $(X,\|\cdot\|')$ is complete. But $\mbox{Id}:(X,\|\cdot\|)\longrightarrow (X,\|\cdot\|')$ is not bounded since $T$ is not bounded by construction. So the two norms are not equivalent.

Note that it is easy to adapt the argument to construct infinitely many complete pairwise non-equivalent norms on $X$. Just because $J$ contains $\mathbb{N}\simeq \mathbb{N}\times \mathbb{N}$.

In a Banach space, a sequence is Cauchy if and only if it converges. So you are asking when two complete norms $\|x\|_1$ and $\|x\|_2$ on a given space $X$ yield the same converging sequences.

Since these are metric spaces, the topologies are sequential. So you are asking when two distinct (complete) norms define the same topology. That is, when $$\mbox{Id}:(X,\|\cdot\|_1)\longrightarrow (X,\|\cdot\|_2)$$ is a homeomorphism. Since this is an invertible linear map, this is equivalent to boundedness in both directions. That is, to the existence of positive constants $A,B$ such that $$A\|x\|_1\leq \|x\|_2\leq B\|x\|_1\quad \forall x\in X.$$ We say that the norms are equivalent.

But by Banach isomorphism theorem (or the open mapping if you prefer), it is sufficient that one side of the inequality hold. In other terms: if every Cauchy sequence for one norm is Cauchy for the other, the norms are equivalent and the converse holds.

Example: on a finite-dimensional real or complex vector space, all the norms are equivalent and complete.

Example: $\|f\|_\infty=\sup_{[0,1]}|f(x)|$ and $\|f\|=\|f\|_\infty+|f(0)|$ are two equivalent complete norms on $C^0([0,1])$.

Oops, I forgot to answer the question...

Counterexample: David Mitra's counterexample is perfect. See this article for the existence of spaces with infinitely many complete norms such that the resulting Banach spaces are pairwise non-isomorphic (in particular, the norms are pairwise non-equivalent, but this result is much stronger).

• The article in your last two lines shows something much stronger than what you summarize. Your example already allows you to produce infinitely many non-equivalent norms (by splitting up your Hamel basis into infinitely many disjoint subsets that serve as $J_0$ in your construction). What you can't guarantee by this procedure is that the resulting Banach spaces are not related by an isomorphism (not necessarily the identity). – Martin Jun 23 '13 at 3:23
• @Martin Sure. I just wanted to give the easiest example (can I say that?) to the question. Actually, I don't have access to the article I linked to... – Julien Jun 23 '13 at 3:41

It's definitely possible. Consider the real numbers with (1) the usual metric and (2) the discrete metric. Both are complete, yet the only Cauchy sequences in (2) are the eventually constant sequences, while (1) has many more Cauchy sequences.

• The discrete metric induces a norm? – David Mitra Jun 21 '13 at 20:54
• Meh. My bad. I was thinking of mere complete metric spaces. Bring on the downvotes! – Greg Martin Jun 22 '13 at 0:36
• @GregMartin This is how I picture this – Pedro Tamaroff Jun 23 '13 at 3:06