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As we known, Banach spaces are normed vector spaces where Cauchy sequences converge. Can someone give me some examples of vector spaces, with a defined norm, which are not complete?

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    $\begingroup$ Compactly supported continuous functions on $\mathbb R$ with respect to sup norm. $\endgroup$
    – user276115
    Commented Sep 30, 2016 at 16:20

5 Answers 5

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As a Functional Analysis example, consider the space $X=C^0([0,1])$, the space of the continuous functions on the interval $[0,1]$. Consider the norm $\|\cdot\|_2$ on $X$ defined by $$ \|f\|_2=\left(\int_0^1|f(t)|^2\, dt\right)^{1/2}. $$ Then $(X,\|\cdot\|_2)$ is not complete. In fact, you can find a $\|\cdot\|_2$-Cauchy sequence which would converge to a discountinuous function (hence to something outside $X$). For example you can approximate (in the sense of the norm $\|\cdot\|_2$) the step function with jump at $1/2$ by menas of continuous functions. This would not be possible in the sense of the norm $\|\cdot\|_\infty$! After all, $(X,\|\cdot\|_\infty)$ is a complete normed space.

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  • $\begingroup$ nice answer, could you give a formal proof ? $\endgroup$ Commented Oct 2, 2016 at 13:01
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    $\begingroup$ @RafaelRojas. Consider the continuous functions $f_n$ equal to 0 in the interval $[0,1/2-2^{-n}]$, equal to 1 in $[1/2+2^{-n},1]$, and linearly interpolated in $[1/2-2^{-n},1/2+2^{-n}]$. It is a direct calculation to prove that $\|f_n-f_m\|_2\leq 2^{-n} $ for any $m\geq n$. So the sequence is Cauchy in $X$. At the same time it is easy to show that $\|f-f_n\|_2\to 0$, where $f\notin X$ is the step function equal to 0 in $[0,1/2]$ and to 1 in $[1/2,1]$. From this it follows that $f_n$ has not limit in $X$: any continuous function has a positive $\|\cdot\|$-distance from $f$. $\endgroup$
    – guestDiego
    Commented Oct 2, 2016 at 14:39
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The rationals $\Bbb Q$ form a field, which is thus a vector space over itself. They inherit a norm from $\Bbb R$, but are not complete with respect to this norm. (Construct a sequence which seems to converge to an irrational number.)

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    $\begingroup$ If you're gonna be pedantic, rationals don't inherent from the reals it's norm but rather the real norm comes from the rationals as it is constructed from the rational norm. $\endgroup$ Commented Oct 1, 2016 at 6:08
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    $\begingroup$ But the vector space of $\mathbb{Q}$ over $\mathbb{Q}$ is not a normed vector space. The definition of normed vector space requires a vector space over $\mathbb{R}$ or $\mathbb{C}$ from the outset. $\endgroup$
    – Imperton
    Commented Jan 8, 2022 at 5:05
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    $\begingroup$ why can't you have a vector space over $\mathbb{Q}$? The rationals are a field. Can you not have vector spaces over any field? $\endgroup$ Commented Feb 3, 2023 at 4:18
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    $\begingroup$ @william_grisaitis Yes, but by definition normed vector spaces are over $\mathbb{R}$ or $\mathbb{C}$, at least that's what every functional analysis textbook I've read says. $\endgroup$
    – Imperton
    Commented Feb 3, 2023 at 4:35
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Consider $V=C([0,1])$, the space of continuous real functions over $[0,1]$ with the norm $$ \|f\|=\int_0^1 |f(x)|\,dx $$ Its completion is the space $L^1([0,1])$, the space of Lebesgue-integrable function modulo the subspace of functions that are $0$ almost everywhere.

It should be easy for you to find a Cauchy sequence in $V$ that does not converge.

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  • $\begingroup$ Sorry for the same choice of $C([0,1])$! $\endgroup$
    – guestDiego
    Commented Sep 30, 2016 at 16:28
  • $\begingroup$ @guestDiego The OP didn't ask for a prehilbert space that's not Hilbert. ;-) $\endgroup$
    – egreg
    Commented Sep 30, 2016 at 16:36
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    $\begingroup$ @OGC No, they aren't continuous on $[0,1]$ $\endgroup$
    – egreg
    Commented Sep 7, 2018 at 21:47
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    $\begingroup$ Sorry, I think it's $f_{n} = x^{n}$. I guess the trick here is to use step functions $\endgroup$
    – OGC
    Commented Sep 7, 2018 at 21:50
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    $\begingroup$ @OGC That's right $\endgroup$
    – egreg
    Commented Sep 7, 2018 at 21:59
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The other two answers are the first that came to my mind as well. You may already know this, but every finite dimensional normed vector space is complete. Hence, any counter-example will have to be something with an infinite dimensional vector space.

there are two basic considerations for vector spaces:

  1. Some Abelian group structure on the vectors $+: V\times V \to V$

  2. Scalar multiplication $(\cdot) : F \times V \to V$.

A norm on a vector space induces a metric topology, and what needs to happen is $\|\cdot\|$ should be continuous, and the aforementioned requirements are functions $(+,(\cdot))$ are continuous as well. Basically, the norm has to agree with the linear structure of the space in question. So, in this sense you can imagine that if you randomly choose a norm space, it is unlikely to be complete. (But also, note that every norm on a finite dimensional vector space induces the same metric topology.)

On the other hand, every norm vector space sits densely inside of a Banach (complete) space with the same norm! This is done by just considering the completion of the vector space. For example, let $\phi([-1,1])$ and $P([-1,1])$ be the space of step functions and polynomials with some $L^p$ norm. Then:

$$\phi([-1,1]) \subseteq P([-1,1]) \subseteq C([-1,1]) \subseteq L^p([-1,1]).$$

In fact, each is a dense subspace of the next, so they all sit densely inside $L^p([-1,1])$.

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    $\begingroup$ So, when a vector space, $V$ is non-Banach, this is because we can have Cauchy sequences that converge to something that's in $W$ where $V$ is a subspace of $W$, right? Its not that we can find some Cauchy sequence that just goes crazy and doesn't converge to anything? $\endgroup$ Commented May 7, 2022 at 21:45
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The space of polynomials over the field of real numbers is not a Banach space. For details, see the following questions. Note that the norms are different.

Proving that $P[0, 1]$, the space of all polynomials on $[0, 1]$ is not complete.

The Space of Polynomials is Incomplete with Respect to the Summed Coeffcient Norm

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