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 $f(x)=\begin{cases} x &\text{ if } x \in (0,1)\\ 1/2 &\text{otherwise} \end{cases}$ $$f$$ 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 $g(x)=\begin{cases} 0 & \text{ if } x \notin \mathbb Q \text{ or if } x = 0\\ \frac{(-1)^q (q-1)}{q} & \text{ if } x = \frac{p}{q} \neq 0 \text{, with } p, q \text{ relatively prime} \end{cases}$ 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.

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!

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 $$x,y \in [0,+\infty)$$
\begin{align*}
\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 $$x \in [0,+\infty)$$
\begin{align*}
\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}
\end{align*}

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