Tag Archives: compacity

Bounded functions and infimum, supremum

According to the extreme value theorem, a continuous real-valued function \(f\) in the closed and bounded interval \([a,b]\) must attain a maximum and a minimum, each at least once.

Let’s see what can happen for non-continuous functions. We consider below maps defined on \([0,1]\).

First let’s look at \[
x &\text{ if } x \in (0,1)\\
1/2 &\text{otherwise}
\end{cases}\] \(f\) is bounded on \([0,1]\), continuous on the interval \((0,1)\) but neither at \(0\) nor at \(1\). The infimum of \(f\) is \(0\), its supremum \(1\), and \(f\) doesn’t attain those values. However, for \(0 < a < b < 1\), \(f\) attains its supremum and infimum on \([a,b]\) as \(f\) is continuous on this interval.

Bounded function that doesn’t attain its infimum and supremum on all \([a,b] \subseteq [0,1]\)

The function \(g\) defined on \([0,1]\) by \[
0 & \text{ if } x \notin \mathbb Q \text{ or if } x = 0\\
\frac{(-1)^q (q-1)}{q} & \text{ if } x = \frac{p}{q} \neq 0 \text{, with } p, q \text{ relatively prime}
\end{cases}\] is bounded, as for \(x \in \mathbb Q \cap [0,1]\) we have \[
\left\vert g(x) \right\vert < 1.\] Hence \(g\) takes values in the interval \([-1,1]\). We prove that the infimum of \(g\) is \(-1\) and its supremum \(1\) on all intervals \([a,b]\) with \(0 < a < b <1\). Consider \(\varepsilon > 0\) and an odd prime \(q\) such that \[
q > \max(\frac{1}{\varepsilon}, \frac{1}{b-a}).\] This is possible as there are infinitely many prime numbers. By the pigeonhole principle and as \(0 < \frac{1}{q} < b-a\), there exists a natural number \(p\) such that \(\frac{p}{q} \in (a,b)\). We have \[ -1 < g \left(\frac{p}{q} \right) = \frac{(-1)^q (q-1)}{q} = - \frac{q-1}{q} <-1 +\varepsilon\] as \(q\) is supposed to be an odd prime with \(q > \frac{1}{\varepsilon}\). This proves that the infimum of \(g\) is \(-1\). By similar arguments, one can prove that the supremum of \(g\) on \([a,b]\) is \(1\).

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.

What about infinite-dimensional spaces?

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 \[
f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\
f_n(\frac{1}{2^n})=1 \\
\end{cases}\] 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!

Counterexamples around Dini’s theorem

In this article we look at counterexamples around Dini’s theorem. Let’s recall:

Dini’s theorem: If \(K\) is a compact topological space, and \((f_n)_{n \in \mathbb N}\) is a monotonically decreasing sequence (meaning \(f_{n+1}(x) \le f_n(x)\) for all \(n \in \mathbb N\) and \(x \in K\)) of continuous real-valued functions on \(K\) which converges pointwise to a continuous function \(f\), then the convergence is uniform.

We look at what happens to the conclusion if we drop some of the hypothesis.

Cases if \(K\) is not compact

We take \(K=(0,1)\), which is not closed equipped with the common distance. The sequence \(f_n(x)=x^n\) of continuous functions decreases pointwise to the always vanishing function. But the convergence is not uniform because for all \(n \in \mathbb N\) \[\sup\limits_{x \in (0,1)} x^n = 1\]

The set \(K=\mathbb R\) is closed but unbounded, hence also not compact. The sequence defined by \[f_n(x)=\begin{cases}
0 & \text{for } x < n\\ \frac{x-n}{n} & \text{for } n \le x < 2n\\ 1 & \text{for } x \ge 2n \end{cases}\] is continuous and monotonically decreasing. It converges to \(0\). However, the convergence is not uniform as for all \(n \in \mathbb N\): \(\sup\{f_n(x) : x \in \mathbb R\} =1\). Continue reading Counterexamples around Dini’s theorem

A counterexample to Krein-Milman theorem

In the theory of functional analysis, the Krein-Milman theorem states that for a separated locally convex topological vector space \(X\), a compact convex subset \(K\) is the closed convex hull of its extreme points.

For the reminder, an extreme point of a convex set \(S\) is a point in \(S\) which does not lie in any open line segment joining two points of S. A point \(p \in S\) is an extreme point of \(S\) if and only if \(S \setminus \{p\}\) is still convex.

In particular, according to the Krein-Milman theorem, a non-empty compact convex set has a non-empty set of extreme points. Let see what happens if we weaken some hypothesis of Krein-Milman theorem. Continue reading A counterexample to Krein-Milman theorem

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

Counterexample around Arzela-Ascoli theorem

Let’s recall Arzelà–Ascoli theorem:

Suppose that \(F\) is a Banach space and \(E\) a compact metric space. A subset \(\mathcal{H}\) of the Banach space \(\mathcal{C}_F(E)\) is relatively compact in the topology induced by the uniform norm if and only if it is equicontinuous and and for all \(x \in E\), the set \(\mathcal{H}(x)=\{f(x) \ | \ f \in \mathcal{H}\}\) is relatively compact.

We look here at what happens if we drop the requirement on space \(E\) to be compact and provide a counterexample where the conclusion of Arzelà–Ascoli theorem doesn’t hold anymore.

We take for \(E\) the real interval \([0,+\infty)\) and for all \(n \in \mathbb{N} \setminus \{0\}\) the real function
\[f_n(t)= \sin \sqrt{t+4 n^2 \pi^2}\] We prove that \((f_n)\) is equicontinuous, converges pointwise to \(0\) but is not relatively compact.

According to the mean value theorem, for all \(x,y \in \mathbb{R}\)
\[\vert \sin x – \sin y \vert \le \vert x – y \vert\] Hence for \(n \ge 1\) and \(x,y \in [0,+\infty)\)
\vert f_n(x)-f_n(y) \vert &\le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{y+4 n^2 \pi^2} \vert \\
&= \frac{\vert x – y \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{y+4 n^2 \pi^2}} \\
&\le \frac{\vert x – y \vert}{4 \pi}
\end{align*} using multiplication by the conjugate.

Which enables to prove that \((f_n)\) is equicontinuous.

We also have for \(n \ge 1\) and \(x \in [0,+\infty)\)
\vert f_n(x) \vert &= \vert f_n(x) – f_n(0) \vert \le \vert \sqrt{x+4 n^2 \pi^2} -\sqrt{4 n^2 \pi^2} \vert \\
&= \frac{\vert x \vert}{\sqrt{x+4 n^2 \pi^2} +\sqrt{4 n^2 \pi^2}} \\
&\le \frac{\vert x \vert}{4 n \pi}

Hence \((f_n)\) converges pointwise to \(0\) and for \(t \in [0,+\infty), \mathcal{H}(t)=\{f_n(t) \ | \ n \in \mathbb{N} \setminus \{0\}\}\) is relatively compact

Finally we prove that \(\mathcal{H}=\{f_n \ | \ n \in \mathbb{N} \setminus \{0\}\}\) is not relatively compact. While \((f_n)\) converges pointwise to \(0\), \((f_n)\) does not converge uniformly to \(f=0\). Actually for \(n \ge 1\) and \(t_n=\frac{\pi^2}{4} + 2n \pi^2\) we have
\[f_n(t_n)= \sin \sqrt{\frac{\pi^2}{4} + 2n \pi^2 +4 n^2 \pi^2}=\sin \sqrt{\left(\frac{\pi}{2} + 2 n \pi\right)^2}=1\] Consequently for all \(n \ge 1\) \(\Vert f_n – f \Vert_\infty \ge 1\). If \(\mathcal{H}\) was relatively compact, \((f_n)\) would have a convergent subsequence with \(f=0\) for limit. And that cannot be as for all \(n \ge 1\) \(\Vert f_n – f \Vert_\infty \ge 1\).

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

A compact convex set whose extreme points set is not close

Let’s remind that an extreme point \(c\) of a convex set \(C\) in a real vector space \(E\) is a point in \(C\) which does not lie in any open line segment joining two points of \(C\).

The specific case of dimension \(2\)

Proposition: when \(C\) is closed and its dimension is equal to \(2\), the set \(\hat{C}\) of its extreme points is closed.
Continue reading A compact convex set whose extreme points set is not close