# A non complete normed vector space

Consider a real normed vector space $$V$$. $$V$$ is called complete if every Cauchy sequence in $$V$$ converges in $$V$$. A complete normed vector space is also called a Banach space.

A finite dimensional vector space is complete. This is a consequence of a theorem stating that all norms on finite dimensional vector spaces are equivalent.

There are many examples of Banach spaces with infinite dimension like $$(\ell_p, \Vert \cdot \Vert_p)$$ the space of real sequences endowed with the norm $$\displaystyle \Vert x \Vert_p = \left( \sum_{i=1}^\infty \vert x_i \vert^p \right)^{1/p}$$ for $$p \ge 1$$, the space $$(C(X), \Vert \cdot \Vert)$$ of real continuous functions on a compact Hausdorff space $$X$$ endowed with the norm $$\displaystyle \Vert f \Vert = \sup\limits_{x \in X} \vert f(x) \vert$$ or the Lebesgue space $$(L^1(\mathbb R), \Vert \cdot \Vert_1)$$ of Lebesgue real integrable functions endowed with the norm $$\displaystyle \Vert f \Vert = \int_{\mathbb R} \vert f(x) \vert \ dx$$.

Let’s give an example of a non complete normed vector space. Let $$(P, \Vert \cdot \Vert_\infty)$$ be the normed vector space of real polynomials endowed with the norm $$\displaystyle \Vert p \Vert_\infty = \sup\limits_{x \in [0,1]} \vert p(x) \vert$$. Consider the sequence of polynomials $$(p_n)$$ defined by
$p_n(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \cdots + \frac{x^n}{2^n} = \sum_{k=0}^{n} \frac{x^k}{2^k}.$ For $$m < n$$ and $$x \in [0,1]$$, we have $\vert p_n(x) - p_m(x) \vert = \left\vert \sum_{i=m+1}^n \frac{x^i}{2^i} \right\vert \le \sum_{i=m+1}^n \frac{1}{2^i} \le \frac{1}{2^m}$ which proves that $$(p_n)$$ is a Cauchy sequence. Also for $$x \in [0,1]$$ $\lim\limits_{n \to \infty} p_n(x) = p(x) \text{ where } p(x) = \frac{1}{1 - \frac{x}{2}}.$ As uniform converge implies pointwise convergence, if $$(p_n)$$ was convergent in $$P$$, it would be towards $$p$$. But $$p$$ is not a polynomial function as none of its $$n$$th-derivative always vanishes. Hence $$(p_n)$$ is a Cauchy sequence that doesn't converge in $$(P, \Vert \cdot \Vert_\infty)$$, proving as desired that this normed vector space is not complete. More generally, a normed vector space with countable dimension is never complete. This can be proven using Baire category theorem which states that a non-empty complete metric space is not the countable union of nowhere-dense closed sets.

# A non-compact closed ball

Consider a normed vector space $$(X, \Vert \cdot \Vert)$$. If $$X$$ is finite-dimensional, then a subset $$Y \subset X$$ is compact if and only if it is closed and bounded. In particular a closed ball $$B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}$$ is always compact if $$X$$ is finite-dimensional.

### The space $$A=C([0,1],\mathbb R)$$

Consider the space $$A=C([0,1],\mathbb R)$$ of the real continuous functions defined on the interval $$[0,1]$$ endowed with the sup norm:
$\Vert f \Vert = \sup\limits_{x \in [0,1]} \vert f(x) \vert$
Is the closed unit ball $$B_1[0]$$ compact? The answer is negative and we provide two proofs.

The first one is based on open covers. For $$n \ge 1$$, we denote by $$f_n$$ the piecewise linear map defined by $\begin{cases} f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\ f_n(\frac{1}{2^n})=1 \\ f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0 \end{cases}$ All the $$f_n$$ belong to $$B_1[0]$$. Moreover for $$1 \le n < m$$ we have $$\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}$$. Hence the supports of the $$f_n$$ are disjoint and $$\Vert f_n – f_m \Vert = 1$$.

Now consider the open cover $$\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}$$. For $$x \in B_1[0]\}$$ and $$u,v \in B_{\frac{1}{2}}(x)$$, $$\Vert u -v \Vert < 1$$. Therefore, each $$B_{\frac{1}{2}}(x)$$ contains at most one $$f_n$$ and a finite subcover of $$\mathcal U$$ will contain only a finite number of $$f_n$$ proving that $$A$$ is not compact.

Second proof based on convergent subsequence. As $$A$$ is a metric space, it is enough to prove that $$A$$ is not sequentially compact. Consider the sequence of functions $$g_n : x \mapsto x^n$$. The sequence is bounded as for all $$n \in \mathbb N$$, $$\Vert g_n \Vert = 1$$. If $$(g_n)$$ would have a convergent subsequence, the subsequence would converge pointwise to the function equal to $$0$$ on $$[0,1)$$ and to $$1$$ at $$1$$. As this function is not continuous, $$(g_n)$$ cannot have a subsequence converging to a map $$g \in A$$.

### Riesz’s theorem

The non-compactness of $$A=C([0,1],\mathbb R)$$ is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space $$X$$ is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of $$A=C([0,1],\mathbb R)$$ is just standard for infinite-dimensional normed vector spaces!

# A homeomorphism of the unit ball having no fixed point

Let’s recall Brouwer fixed-point theorem.

Theorem (Brouwer): Every continuous function from a convex compact subset $$K$$ of a Euclidean space to $$K$$ itself has a fixed point.

We here describe an example of a homeomorphism of the unit ball of a Hilbert space having no fixed point. Let $$E$$ be a separable Hilbert space with $$(e_n)_{n \in \mathbb{Z}}$$ as a Hilbert basis. $$B$$ and $$S$$ are respectively $$E$$ closed unit ball and unit sphere.

There is a unique linear map $$u : E \to E$$ for which $$u(e_n)=e_{n+1}$$ for all $$n \in \mathbb{Z}$$. For $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$ we have $$u(x)= \sum_{n \in \mathbb{Z}} \xi_n e_{n+1}$$. $$u$$ is isometric as $\Vert u(x) \Vert^2 = \sum_{n \in \mathbb{Z}} \vert \xi_n \vert^2 = \Vert x \Vert^2$ hence one-to-one. $$u$$ is also onto as for $$x = \sum_{n \in \mathbb{Z}} \xi_n e_n \in E$$, $$\sum_{n \in \mathbb{Z}} \xi_n e_{n-1} \in E$$ is an inverse image of $$x$$. Finally $$u$$ is an homeomorphism. Continue reading A homeomorphism of the unit ball having no fixed point

# A solution of a differential equation not exploding in finite time

In this post, I mention that Peano existence theorem is valid for finite dimensional vector spaces, but not for Banach spaces of infinite dimension. I highlight here a second property of ordinary differential equations which is valid for finite dimensional vector spaces but not for infinite dimensional Banach spaces. Continue reading A solution of a differential equation not exploding in finite time

# An unbounded convex not containing a ray

We consider a normed vector space $$E$$ over the field of the reals $$\mathbb{R}$$ and a convex subset $$C \subset E$$.

We suppose that $$0 \in C$$ and that $$C$$ is unbounded, i.e. there exists points in $$C$$ at distance as big as we wish from $$0$$.

The following question arises: “does $$C$$ contains a ray?”. It turns out that the answer depends on the dimension of the space $$E$$. If $$E$$ is of finite dimension, then $$C$$ always contains a ray, while if $$E$$ is of infinite dimension $$C$$ may not contain a ray. Continue reading An unbounded convex not containing a ray

# A compact whose convex hull is not compact

We consider a topological vector space $$E$$ over the field of the reals $$\mathbb{R}$$. The convex hull of a subset $$X \subset E$$ is the smallest convex set that contains $$X$$.

The convex hull may also be defined as the intersection of all convex sets containing X or as the set of all convex combinations of points in X.

The convex hull of $$X$$ is written as $$\mbox{Conv}(X)$$. Continue reading A compact whose convex hull is not compact