Example of a sequence that is Cauchy in a stronger norm and convergent in a weaker norm, but not convergent in the stronger norm?

Question

A norm $\|\cdot\|_1$ on a normed vector space is called stronger than $\|\cdot\|_2$ when $\|x\|_2\leq M\|x\|_1$ for some $M>0$ and all $x$. It is a standard trick (e.g. in proving completeness) to find the limit in a weaker sense first, and then prove that it is a stronger limit, e.g. Cauchy in Norm and Weakly converge Implies Norm convergent. But it always uses some extra information beyond one convergence being weaker than the other. Examples in Cauchy sequence converge for one metric while not converging for another? are not for norm metrics, hence the question.

Answer 1

No ZFC is needed since the completness of any of the norms is not required in OP. Therefore an example can be constructed explicitly.

Let $\mathcal{F}(\mathbb{N})$ denote the space of sequences with finitely many nonzero terms with the standard basis $\{\delta_n\}_{n=1}^\infty.$ Consider two norms $$\|x\|_1=\sum_{n=1}^\infty |x_n|,\quad\|x\|_2=\|x\|_1+\left |x_1-\sum_{n=2}^\infty nx_n \right |$$ Then $\|\cdot \|_1\le \|\cdot \|_2.$ The sequence $v_k=\delta_1-{1\over k}\delta_k$ tends to $\delta_1$ with respect to $\|\cdot \|_1$ norm. On the other hand $\|v_k-\delta_1\|_2={1\over k}+1$ for $k\ge 2.$ Hence the sequence $v_k$ is not convergent to $\delta_1$ with respect to $\|\cdot \|_2$ norm. However for $k,l\ge 2$ and $k\neq l$ we have $$\|v_k-v_l\|_2=\left \|{1\over l}\delta_l-{1\over k}\delta_k\right \|_2={1\over l}+{1\over k}$$ therefore $v_k$ satisfies the Cauchy condition with respect to $\|\cdot \|_2.$ The sequence $v_k$ is not convergent with respect to $\|\cdot\|_2$ as the only possible limit would be equal $\delta_1.$

What follows is based on a beautiful idea of @SeverinSchraven, who kindly agreed to include his results into the answer and took care of the re-edition.

Added: The convergence with respect to $\Vert \cdot \Vert_2$ is a bit difficult to grasp and one might wonder what the completion looks like.

The example shows that this norm keeps track of some mass that "escapes to infinity". It is a bit hard to imagine how this exactly works, thus we change the norm to something that we are a more familiar with at the expense of making the underlying space more complicated. Namely, consider the following map

$$ i: \mathcal{F}(\mathbb{N}) \rightarrow \ell^1(\mathbb{N} \cup \{ \infty\}), (x_n)_{n\in \mathbb{N}} \mapsto ((x_n)_{n\in \mathbb{N}}, x_1 - \sum_{n=2}^\infty n x_n). $$

The first coordinate keeps track of $\Vert \cdot \Vert_1$ and the second coordinate takes care of the mass wandering off to infinity. We define $$ \Vert ((x_n)_{n\in \mathbb{N}}, x_\infty)\Vert_{\ell^1} = \Vert (x_n)_{n\in \mathbb{N}} \Vert_1 + \vert x_\infty \vert, $$ which is just the usual $\ell^1$ norm. One readily checks that $i: (\mathcal{F}(\mathbb{N}), \Vert \cdot \Vert_1) \rightarrow (\ell^1(\mathbb{N}\cup \{\infty\}, \Vert \cdot \Vert_{\ell^1})$ is an isometry. Thus, $\mathcal{F}(\mathbb{N})$ is isomorphic to $i(\mathcal{F}(\mathbb{N}))$ as a normed space. Hence, if we want to classify the completion of $\mathcal{F}(\mathbb{N})$ (up to isomorphism), then it is enough to find the completion of $i(\mathcal{F}(\mathbb{N}))$.

We are in much better shape now as $(\ell^1(\mathbb{N}\cup \{\infty\}), \Vert \cdot \Vert_{\ell^1})$ is a Banach space! Thus, the completion of $i(\mathcal{F}(\mathbb{N}))$ is (isomorphic to) the closure of $i(\mathcal{F}(\mathbb{N}))$ in $(\ell^1(\mathbb{N}\cup \{\infty\}), \Vert \cdot \Vert_{\ell^1})$.

We claim that the closure of $i(\mathcal{F}(\mathbb{N}))$ in $\ell^1(\mathbb{N}\cup \{\infty\})$ is just $\ell^1(\mathbb{N}\cup \{\infty\})$ itself. For this it is enough to show that $\mathcal{F}(\mathbb{N}\cup \{\infty\}) \subseteq \overline{i(\mathcal{F}(\mathbb{N}))}$ as the former is dense in $\ell^1(\mathbb{N}\cup \{\infty\})$ (just truncate). Let's fix any $((x_1, x_2, \dots, x_N,0, 0, 0, \dots), x_\infty)\in \mathcal{F}(\mathbb{N}\cup \{\infty\})$ and set $x=(x_1, x_2, \dots, x_N,0, 0, 0, \dots)$, then we can pick for $M>N$

\begin{align*} v_M &= i(x +\frac{1}{M}(-x_\infty+x_1- \sum_{n=2}^N n x_n) \delta_M) \\ &= (x+ \frac{1}{M}(-x_\infty+x_1- \sum_{n=2}^N n x_n) \delta_M, x_\infty). \end{align*}

Now we need to show that $(v_M)_{M\geq N+1}$ converges to $(x, x_\infty)$ in $\ell^1(\mathbb{N}\cup \{\infty\})$. We have

\begin{align*} \Vert v_M - (x, x_\infty) \Vert_{\ell^1} = \frac{1}{M}\left\vert -x_\infty+x_1- \sum_{n=2}^N n x_n\right\vert \end{align*}

which clearly tends to zero.

Note that we actually have $\ell^1(\mathbb{N}\cup \{\infty\}) \cong \ell^1(\mathbb{N})$ (as Banach spaces) via the isomorphism $$ \ell^1(\mathbb{N}\cup \{\infty\}) \rightarrow \ell^1(\mathbb{N}), ((x_n)_{n\in \mathbb{N}}, x_\infty) \mapsto (x_\infty, x_1, x_2, x_3, \dots). $$

Answer 2

Using a Hamel basis of an infinite dimensional Banach space $(X,\|\cdot\|_2)$ one finds a discontinuous linear functional $f$. We define a strictly stronger norm $\|x\|_1=\|x\|_2+|f(x)|$. The open mapping theorem implies that $(X,\|\cdot\|_1)$ is not complete but every $\|\cdot\|_1$-Cauchy-sequence is also $\|\cdot\|_2$-Cauchy and hence $\|\cdot\|_2$-convergent.