Example of a non complete normed vector space.

Question

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?

Answer 1

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.

Answer 2

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.)

Answer 3

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.

Answer 4

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 $L^p([-1,1])$.

Answer 5

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