Does there exist a normed vector space $(X,\|\cdot\|)$ over $\mathbb R$ or $\mathbb C$ such that

  • the metric induced by the norm $(x,y)\mapsto\|x-y\|$ is not complete; but
  • there exists some other metric $d$ on $X$ that is complete and defines the same topology on $X$ as the norm?

It may be possible, since completeness of a metric is not a topological property, but I can't think of an obvious counterexample.

In other words: is every completely metrizable (by some metric) normed vector space necessarily a Banach space (i.e., complete also in the norm metric)?

  • 2
    $\begingroup$ An example would have to have uncountable linear dimension (by Baire). $\endgroup$ Oct 27, 2014 at 5:08
  • $\begingroup$ @JonasMeyer That's true. A finite-dimensional normed vector space is always Banach, so $X$ should be infinite-dimensional. In turn, one can use the Baire category theorem to show that a normed vector space of countably infinite dimension is not only not Banach, but not complete$\textit{ly metrizable}$, either. So $\operatorname{dim}X>\#\mathbb N$, indeed. $\endgroup$
    – triple_sec
    Oct 27, 2014 at 5:10

2 Answers 2


The answer is yes, any such space must be a Banach space. This result was proved by Victor Klee in 1952 and answered a question first asked by Banach in 1932.

V. L. Klee, Invariant metrics in groups (Solution of a problem of Banach), Proc. Amer. Math. Soc., 3 (1952), 484–487.

  • $\begingroup$ Yes, thank you for pointing it out. Actually, I had found this paper earlier, but I'm not very comfortable with group theory, I tried to “translate” its main arguments to functional-analytic language in my answer infra. Thank you for providing this reference. $\endgroup$
    – triple_sec
    Oct 27, 2014 at 9:11

I think I got it. There is no such counterexample. I welcome and appreciate any comments on the proof below, which hinges on ideas presented by Klee (1952).

$\textbf{Claim:}\quad$If $(X,\|\cdot\|)$ is a completely metrizable normed vector space, then it is a Banach space.

Proof:$\quad$Suppose that $(X,\|\cdot\|)$ is a normed vector space and $d$ is such a complete metric on it that defines the same topology as $\|\cdot\|$. Consider the double dual $X^{**}$, the vector space of bounded linear functionals on the space of bounded linear functionals on $X$. Let $\psi:X\to X^{**}$ denote the “natural map” (defined such that if $f\in X^*$, then $\psi(x)(f)\equiv f(x)$ for any $x\in X$). One can use the Hahn–Banach theorem to show that $\psi$ is an injective linear isometry between $(X,\|\cdot\|)$ and $(X^{**},\|\cdot\|^{**})$, where $\|\cdot\|^{**}$ is the operator norm on $X^{**}$.

Let $Y$ denote the closure of $\psi(X)$ in $X^{**}$. Since $(X^{**},\|\cdot\|^{**})$ is always a Banach space, it follows that $(Y,\|\cdot\|^{**})$ is a Banach space in its own right as a closed subspace.

Define a metric $\delta$ on $\psi(X)$ as follows: $$\delta(\psi(x),\psi(y))\equiv d(x,y)\quad\forall x,y\in X.$$ It is not difficult to show that $\delta$ is a complete metric on $\psi(X)$, given that $d$ is complete on $X$. Moreover, $\delta$ defines the same topology on $\psi(X)$ as the relative $\|\cdot\|^{**}$-norm topology, given that any $\delta$-open ball in $\psi(X)$ contains a $\|\cdot\|^{**}$-open ball and vice versa. (To prove this, one can use the facts that $\psi$ is a linear isometry between $(X,\|\cdot\|)$ and $(\psi(X),\|\cdot\|^{**})$ and that any $d$-open subset of $X$ is also $\|\cdot\|$-open.) Hence, $\psi(X)$ is a completely metrizable subspace of $Y$, so $\psi(X)$ must be a $G_{\delta}$ subset of $Y$ (a countable intersection of $\|\cdot\|^{**}$-open subsets of $Y$): $$\psi(X)=\bigcap_{n=1}^{\infty} U_n,$$ where, for each $n\in\mathbb N$, $U_n\subseteq Y$ is $\|\cdot\|^{**}$-open (see, for example, Lemma 3.33 in Aliprantis–Border, 2006, p. 88 or Theorem 24.12 in Willard, 2004, pp. 179–180). These open sets are also dense in $Y$ given that $\psi(X)$ is, so $$Z\equiv Y\setminus\psi(X)=\bigcup_{n=1}^{\infty}Y\setminus U_n$$ is of the first category, since $Y\setminus U_n$ is closed and nowhere dense for each $n\in\mathbb N$.

Suppose that $Z$ is not empty. Then, there exists some $z\in Z$. Let $$W\equiv\{z+\psi(x)\,|\,x\in X\}.$$ Since $\psi(X)$ is a subspace of $Y$, it is easy to see that $W\subseteq Z$ and hence $W$ is of the first category. It follows from a simple translation argument that $\psi(X)$ is also of the first category. Consequently, $Y=\psi(X)\cup Z$ is of the first category, too, contradicting Baire's category theorem ($Y$ is a Banach space). Hence, $Z$ must be empty. In other words, $\psi(X)=Y$ is a Banach space with the norm $\|\cdot\|^{**}$, and since $\psi$ is a linear isometry between $(X,\|\cdot\|)$ and $(\psi(X),\|\cdot\|^{**})$, $(X,\|\cdot\|)$ is a Banach space, too. The proof is complete. $\blacksquare$


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