Tag Archives: maps

Analysis

Counterexamples around Fubini’s theorem

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We present here some counterexamples around the Fubini theorem.

We recall Fubini’s theorem for integrable functions:
let \(X\) and \(Y\) be \(\sigma\)-finite measure spaces and suppose that \(X \times Y\) is given the product measure. Let \(f\) be a measurable function for the product measure. Then if \(f\) is \(X \times Y\) integrable, which means that \(\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) < \infty\), we have \[\int_X \left( \int_Y f(x,y) dy \right) dx = \int_Y \left( \int_X f(x,y) dx \right) dy = \int_{X \times Y} f(x,y) d(x,y)\] Let's see what happens when some hypothesis of Fubini's theorem are not fulfilled. Continue reading Counterexamples around Fubini’s theorem →

Analysis

A discontinuous real convex function

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Consider a function \(f\) defined on a real interval \(I \subset \mathbb R\). \(f\) is called convex if: \[\forall x, y \in I \ \forall \lambda \in [0,1]: \ f((1-\lambda)x+\lambda y) \le (1-\lambda) f(x) + \lambda f(y)\]

Suppose that \(I\) is a closed interval: \(I=[a,b]\) with \(a < b\). For \(a < s < t < u < b\) one can prove that: \[\frac{f(t)-f(s)}{t-s}\le \frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.\] It follows from those relations that \(f\) has left-hand and right-hand derivatives at each point of the interior of \(I\). And therefore that \(f\) is continuous at each point of the interior of \(I\).
Is a convex function defined on an interval \(I\) continuous at all points of the interval? That might not be the case and a simple example is the function: \[\begin{array}{l|rcl}
f : & [0,1] & \longrightarrow & \mathbb R \\
& x & \longmapsto & 0 \text{ for } x \in (0,1) \\
& x & \longmapsto & 1 \text{ else}\end{array}\]

It can be easily verified that \(f\) is convex. However, \(f\) is not continuous at \(0\) and \(1\).

Topology

Counterexamples to Banach fixed-point theorem

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Let \((X,d)\) be a metric space. Then a map \(T : X \to X\) is called a contraction map if it exists \(0 \le k < 1\) such that \[d(T(x),T(y)) \le k d(x,y)\] for all \(x,y \in X\). According to Banach fixed-point theorem, if \((X,d)\) is a complete metric space and \(T\) a contraction map, then \(T\) admits a fixed-point \(x^* \in X\), i.e. \(T(x^*)=x^*\).

We look here at counterexamples to the Banach fixed-point theorem when some hypothesis are not fulfilled.

First, let’s consider \[\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
& x & \longmapsto & x+1 \end{array}\] For all \(x,y \in \mathbb R\) we have \(\vert f(x)-f(y) \vert = \vert x- y \vert\). \(f\) is not a contraction, but an isometry. Obviously, \(f\) has no fixed-point.

We now prove that a map satisfying \[d(g(x),g(y)) < d(x,y)\] might also not have a fixed-point. A counterexample is the following map \[\begin{array}{l|rcl} g : & [0,+\infty) & \longrightarrow & [0,+\infty) \\ & x & \longmapsto & \sqrt{1+x^2} \end{array}\] Since \[g^\prime(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for all } \xi \in [0, +\infty),\] by the mean value theorem \[\vert g(x) - g(y)| = \vert g^\prime(\xi)\vert |x-y| < |x-y| \text{ for all } x, y \in [0, +\infty).\] However \(g\) has no fixed-point. Finally, let's have a look to a space \((X,d)\) which is not complete. We take \(a,b \in \mathbb R\) with \(0 < a < 1\) and for \((X,d)\) the space \(X = \mathbb R \setminus \{\frac{b}{1-a}\}\) equipped with absolute value distance. \(X\) is not complete. Consider the map \[\begin{array}{l|rcl} h : & X & \longrightarrow & X \\ & x & \longmapsto & ax + b \end{array}\] \(h\) is well defined as for \(x \neq \frac{b}{1-a}\), \(h(x) \neq \frac{b}{1-a}\). \(h\) is a contraction map as for \(x,y \in \mathbb R\) \[\vert h(x)-h(y) \vert = a \vert x - y \vert \] However, \(h\) doesn't have a fixed-point in \(X\) as \(\frac{b}{1-a}\) is the only real for which \(h(x)=x\).

Analysis

Counterexamples on function limits (part 1)

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Let \(f\) and \(g\) be two real functions and \(a \in \mathbb R \cup \{+\infty\}\). We provide here examples and counterexamples regarding the limits of \(f\) and \(g\).

If \(f\) has a limit as \(x\) tends to \(a\) then \(\vert f \vert\) also?

This is true. It is a consequence of the reverse triangle inequality \[\left\vert \vert f(x) \vert – \vert l \vert \right\vert \le \vert f(x) – l \vert\] Hence if \(\displaystyle \lim\limits_{x \to a} f(x) = l\), \(\displaystyle \lim\limits_{x \to a} \vert f(x) \vert = \vert l \vert\)

Is the converse of previous statement also true?

It is not. Consider the function defined by: \[\begin{array}{l|rcl}
f : & \mathbb R & \longrightarrow & \mathbb R \\
& \frac{1}{n} & \longmapsto & -1 \text{ for } n \ge 1 \text{ integer} \\
& x & \longmapsto & 1 \text{ otherwise} \end{array}\] \(\vert f \vert\) is the constant function equal to \(1\), hence \(\vert f \vert\) has \(1\) for limit as \(x\) tends to zero. However \(\lim\limits_{x \to 0} f(x)\) doesn’t exist. Continue reading Counterexamples on function limits (part 1) →

Topology

A homeomorphism of the unit ball having no fixed point

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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 →

Analysis

A function whose derivative at 0 is one but which is not increasing near 0

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From the mean value theorem, a real function whose derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function \(f\) defined in a non-degenerate interval \(I\) with a strictly positive derivative at a point \(a\) of the interval is strictly increasing near that point. For the proof, we just have to notice that as \(f^\prime\) is continuous and \(f^\prime(a) > 0\), \(f^\prime\) is strictly positive within an interval \(J \subset I\) containing \(a\). By the mean value theorem, \(f\) is strictly increasing on \(J\).

We now suppose that \(f\) is differentiable on an interval \(I\) containing \(0\) with \(f^\prime(0)>0\). For \(x>0\) sufficiently close to zero we have \(\displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0\), hence \(f(x)>f(0)\). But that doesn’t imply that \(f\) is strictly increasing in a neighborhood of zero. Let’s prove it with a counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0 →

Analysis

A decreasing function converging to zero whose derivative diverges (part2)

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In that article, I gave examples of real valued functions defined on \((0,+\infty)\) that converge to zero and whose derivatives diverge. But those functions were not monotonic. Here I give an example of a decreasing real valued function \(g\) converging to zero at \(+\infty\) and whose derivative is unbounded.

We first consider the polynomial map:
\[P(x)=(1+2x)(1-x)^2=1-3x^2+2x^3\] on the segment \(I=[0,1]\). \(P\) derivative equals \(P^\prime(x)=-6x(1-x)\). Therefore \(P\) is decreasing on \(I\). Moreover we have \(P(0)=1\), \(P(1)=P^\prime(0)=P^\prime(1)=0\) and \(P^\prime(1/2)=-3/2\). Continue reading A decreasing function converging to zero whose derivative diverges (part2) →

Analysis

Differentiable functions converging to zero whose derivatives diverge (part1)

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In this article, I consider real valued functions \(f\) defined on \((0,+\infty)\) that converge to zero, i.e.:
\[\lim\limits_{x \to +\infty} f(x) = 0\] If \(f\) is differentiable what can be the behavior of its derivative as \(x\) approaches \(+\infty\)?

Let’s consider a first example:
\[\begin{array}{l|rcl}
f_1 : & (0,+\infty) & \longrightarrow & \mathbb{R} \\
& x & \longmapsto & \frac{1}{x} \end{array}\] \(f_1\) derivative is \(f_1^\prime(x)=-\frac{1}{x^2}\) and we also have \(\lim\limits_{x \to +\infty} f_1^\prime(x) = 0\). Let’s consider more sophisticated cases! Continue reading Differentiable functions converging to zero whose derivatives diverge (part1) →

Topology

Continuous maps that are not closed or not open

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We recall some definitions on open and closed maps. In topology an open map is a function between two topological spaces which maps open sets to open sets. Likewise, a closed map is a function which maps closed sets to closed sets.

For a continuous function \(f: X \mapsto Y\), the preimage \(f^{-1}(V)\) of every open set \(V \subseteq Y\) is an open set which is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in \(Y\) are closed in \(X\). However, a continuous function might not be an open map or a closed map as we prove in following counterexamples. Continue reading Continuous maps that are not closed or not open →