# A differentiable real function with unbounded derivative around zero

Consider the real function defined on $$\mathbb R$$$f(x)=\begin{cases} 0 &\text{for } x = 0\\ x^2 \sin \frac{1}{x^2} &\text{for } x \neq 0 \end{cases}$

$$f$$ is continuous and differentiable on $$\mathbb R\setminus \{0\}$$. For $$x \in \mathbb R$$ we have $$\vert f(x) \vert \le x^2$$, which implies that $$f$$ is continuous at $$0$$. Also $\left\vert \frac{f(x)-f(0)}{x} \right\vert = \left\vert x \sin \frac{1}{x^2} \right\vert \le \vert x \vert$ proving that $$f$$ is differentiable at zero with $$f^\prime(0) = 0$$. The derivative of $$f$$ for $$x \neq 0$$ is $f^\prime(x) = \underbrace{2x \sin \frac{1}{x^2}}_{=g(x)}-\underbrace{\frac{2}{x} \cos \frac{1}{x^2}}_{=h(x)}$ On the interval $$(-1,1)$$, $$g(x)$$ is bounded by $$2$$. However, for $$a_k=\frac{1}{\sqrt{k \pi}}$$ with $$k \in \mathbb N$$ we have $$h(a_k)=2 \sqrt{k \pi} (-1)^k$$ which is unbounded while $$\lim\limits_{k \to \infty} a_k = 0$$. Therefore $$f^\prime$$ is unbounded in all neighborhood of the origin.

# A Riemann-integrable map that is not regulated

For a Banach space $$X$$, a function $$f : [a,b] \to X$$ is said to be regulated if there exists a sequence of step functions $$\varphi_n : [a,b] \to X$$ converging uniformly to $$f$$.

One can prove that a regulated function $$f : [a,b] \to X$$ is Riemann-integrable. Is the converse true? The answer is negative and we provide below an example of a Riemann-integrable real function that is not regulated. Let’s first prove following theorem.

THEOREM A bounded function $$f : [a,b] \to \mathbb R$$ that is (Riemann) integrable on all intervals $$[c, b]$$ with $$a < c < b$$ is integrable on $$[a,b]$$.

PROOF Take $$M > 0$$ such that for all $$x \in [a,b]$$ we have $$\vert f(x) \vert < M$$. For $$\epsilon > 0$$, denote $$c = \inf(a + \frac{\epsilon}{4M},b + \frac{b-a}{2})$$. As $$f$$ is supposed to be integrable on $$[c,b]$$, one can find a partition $$P$$: $$c=x_1 < x_2 < \dots < x_n =b$$ such that $$0 \le U(f,P) - L(f,P) < \frac{\epsilon}{2}$$ where $$L(f,P),U(f,P)$$ are the lower and upper Darboux sums. For the partition $$P^\prime$$: $$a= x_0 < c=x_1 < x_2 < \dots < x_n =b$$, we have \begin{aligned} 0 \le U(f,P^\prime) - L(f,P^\prime) &\le 2M(c-a) + \left(U(f,P) - L(f,P)\right)\\ &< 2M \frac{\epsilon}{4M} + \frac{\epsilon}{2} = \epsilon \end{aligned} We now prove that the function $$f : [0,1] \to [0,1]$$ defined by $f(x)=\begin{cases} 1 &\text{ if } x \in \{2^{-k} \ ; \ k \in \mathbb N\}\\ 0 &\text{otherwise} \end{cases}$ is Riemann-integrable (that follows from above theorem) and not regulated. Let's prove it. If $$f$$ was regulated, there would exist a step function $$g$$ such that $$\vert f(x)-g(x) \vert < \frac{1}{3}$$ for all $$x \in [0,1]$$. If $$0=x_0 < x_1 < \dots < x_n=1$$ is a partition associated to $$g$$ and $$c_1$$ the value of $$g$$ on the interval $$(0,x_1)$$, we must have $$\vert 1-c_1 \vert < \frac{1}{3}$$ as $$f$$ takes (an infinite number of times) the value $$1$$ on $$(0,x_1)$$. But $$f$$ also takes (an infinite number of times) the value $$0$$ on $$(0,x_1)$$. Hence we must have $$\vert c_1 \vert < \frac{1}{3}$$. We get a contradiction as those two inequalities are not compatible.

# A discontinuous midpoint convex function

Let’s recall that a real function $$f: \mathbb R \to \mathbb R$$ is called convex if for all $$x, y \in \mathbb R$$ and $$\lambda \in [0,1]$$ we have $f((1- \lambda) x + \lambda y) \le (1- \lambda) f(x) + \lambda f(y)$ $$f$$ is called midpoint convex if for all $$x, y \in \mathbb R$$ $f \left(\frac{x+y}{2}\right) \le \frac{f(x)+f(y)}{2}$ One can prove that a continuous midpoint convex function is convex. Sierpinski proved the stronger theorem, that a real-valued Lebesgue measurable function that is midpoint convex will be convex.

Can one find a discontinuous midpoint convex function? The answer is positive but requires the Axiom of Choice. Why? Because Robert M. Solovay constructed a model of Zermelo-Fraenkel set theory (ZF), exclusive of the axiom of choice where all functions are Lebesgue measurable. Hence convex according to Sierpinski theorem. And one knows that convex functions defined on $$\mathbb R$$ are continuous.

Referring to my previous article on the existence of discontinuous additive map, let’s use a Hamel basis $$\mathcal B = (b_i)_{i \in I}$$ of $$\mathbb R$$ considered as a vector space on $$\mathbb Q$$. Take $$i_1 \in I$$, define $$f(i_1)=1$$ and $$f(i)=0$$ for $$i \in I\setminus \{i_1\}$$ and extend $$f$$ linearly on $$\mathbb R$$. $$f$$ is midpoint convex as it is linear. As the image of $$\mathbb R$$ under $$f$$ is $$\mathbb Q$$, $$f$$ is discontinuous as explained in the discontinuous additive map counterexample.

Moreover, $$f$$ is unbounded on all open real subsets. By linearity, it is sufficient to prove that $$f$$ is unbounded around $$0$$. Let’s consider $$i_1 \neq i_2 \in I$$. $$G= b_{i_1} \mathbb Z + b_{i_2} \mathbb Z$$ is a proper subgroup of the additive $$\mathbb R$$ group. Hence $$G$$ is either dense of discrete. It cannot be discrete as the set of vectors $$\{b_1,b_2\}$$ is linearly independent. Hence $$G$$ is dense in $$\mathbb R$$. Therefore, one can find a non vanishing sequence $$(x_n)_{n \in \mathbb N}=(q_n^1 b_{i_1} + q_n^2 b_{i_2})_{n \in \mathbb N}$$ (with $$(q_n^1,q_n^2) \in \mathbb Q^2$$ for all $$n \in \mathbb N$$) converging to $$0$$. As $$\{b_1,b_2\}$$ is linearly independent, this implies $$\vert q_n^1 \vert, \vert q_n^2 \vert \underset{n\to+\infty}{\longrightarrow} \infty$$ and therefore $\lim\limits_{n \to \infty} \vert f(x_n) \vert = \lim\limits_{n \to \infty} \vert f(q_n^1 b_{i_1} + q_n^2 b_{i_2}) \vert = \lim\limits_{n \to \infty} \vert q_n^1 \vert = \infty.$

# A positive smooth function with all derivatives vanishing at zero

Let’s consider the set $$\mathcal C^\infty(\mathbb R)$$ of real smooth functions, i.e. functions that have derivatives of all orders on $$\mathbb R$$.

Does a positive function $$f \in \mathcal C^\infty(\mathbb R)$$ with all derivatives vanishing at zero exists?

Such a map $$f$$ cannot be expandable in power series around zero, as it would vanish in a neighborhood of zero. However, the answer to our question is positive and we’ll prove that $f(x) = \left\{\begin{array}{lll} e^{-\frac{1}{x^2}} &\text{if} &x \neq 0\\ 0 &\text{if} &x = 0 \end{array}\right.$ provides an example.

$$f$$ is well defined and positive for $$x \neq 0$$. As $$\lim\limits_{x \to 0} -\frac{1}{x^2} = -\infty$$, we get $$\lim\limits_{x \to 0} f(x) = 0$$ proving that $$f$$ is continuous on $$\mathbb R$$. Let’s prove by induction that for $$x \neq 0$$ and $$n \in \mathbb N$$, $$f^{(n)}(x)$$ can be written as $f^{(n)}(x) = \frac{P_n(x)}{x^{3n}}e^{-\frac{1}{x^2}}$ where $$P_n$$ is a polynomial function. The statement is satisfied for $$n = 1$$ as $$f^\prime(x) = \frac{2}{x^3}e^{-\frac{1}{x^2}}$$. Suppose that the statement is true for $$n$$ then $f^{(n+1)}(x)=\left[\frac{P_n^\prime(x)}{x^{3n}} – \frac{3n P_n(x)}{x^{3n+1}}+\frac{2 P_n(x)}{x^{3n+3}}\right] e^{-\frac{1}{x^2}}$ hence the statement is also true for $$n+1$$ by taking $$P_{n+1}(x)= x^3 P_n^\prime(x) – 3n x^2 P_n(x) + 2 P_n(x)$$. Which concludes our induction proof.

Finally, we have to prove that for all $$n \in \mathbb N$$, $$\lim\limits_{x \to 0} f^{(n)}(x) = 0$$. For that, we use the power expansion of the exponential map $$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$$. For $$x \neq 0$$, we have $\left\vert x \right\vert^{3n} e^{\frac{1}{x^2}} \ge \frac{\vert x \vert^{3n}}{(2n)! \vert x \vert ^{4n}} = \frac{1}{(2n)! \vert x \vert^n}$ Therefore $$\lim\limits_{x \to 0} \left\vert x \right\vert^{3n} e^{\frac{1}{x^2}} = \infty$$ and $$\lim\limits_{x \to 0} f^{(n)}(x) = 0$$ as $$f^{(n)}(x) = \frac{P_n(x)}{x^{3n} e^{\frac{1}{x^2}}}$$ with $$P_n$$ a polynomial function.

# Counterexample around L’Hôpital’s rule

Let us consider two differentiable functions $$f$$ and $$g$$ defined in an open interval $$(a,b)$$, where $$b$$ might be $$\infty$$. If
$\lim\limits_{x \to b^-} f(x) = \lim\limits_{x \to b^-} g(x) = \infty$ and if $$g^\prime(x) \neq 0$$ in some interval $$(c,b)$$, then a version of l’Hôpital’s rule states that $$\lim\limits_{x \to b^-} \frac{f^\prime(x)}{g^\prime(x)} = L$$ implies $$\lim\limits_{x \to b^-} \frac{f(x)}{g(x)} = L$$.

We provide a counterexample when $$g^\prime$$ vanishes in all neighborhood of $$b$$. The counterexample is due to the Austrian mathematician Otto Stolz.

We take $$(0,\infty)$$ for the interval $$(a,b)$$ and $\begin{cases} f(x) &= x + \cos x \sin x\\ g(x) &= e^{\sin x}(x + \cos x \sin x) \end{cases}$ which derivatives are $\begin{cases} f^\prime(x) &= 2 \cos^2 x\\ g^\prime(x) &= e^{\sin x} \cos x (x + \cos x \sin x + 2 \cos x) \end{cases}$ We have $\lim\limits_{x \to \infty} \frac{f^\prime(x)}{g^\prime(x)} = \lim\limits_{x \to \infty} \frac{2 \cos x}{e^{\sin x} (x + \cos x \sin x + 2 \cos x)} = 0,$ however $\frac{f(x)}{g(x)} = \frac{1}{e^{\sin x}}$ doesn’t have any limit at $$\infty$$ as it oscillates between $$\frac{1}{e}$$ and $$e$$.

# On limit at infinity of functions and their derivatives

We consider continuously differentiable real functions defined on $$(0,\infty)$$ and the limits $\lim\limits_{x \to \infty} f(x) \text{ and } \lim\limits_{x \to \infty} f^\prime(x).$

### A map $$f$$ such that $$\lim\limits_{x \to \infty} f(x) = \infty$$ and $$\lim\limits_{x \to \infty} f^\prime(x) = 0$$

Consider the map $$f : x \mapsto \sqrt{x}$$. It is clear that $$\lim\limits_{x \to \infty} f(x) = \infty$$. As $$f^\prime(x) = \frac{1}{2 \sqrt{x}}$$, we have as announced $$\lim\limits_{x \to \infty} f^\prime(x) = 0$$

### A bounded map $$g$$ having no limit at infinity such that $$\lim\limits_{x \to \infty} g^\prime(x) = 0$$

One idea is to take an oscillating map whose wavelength is increasing to $$\infty$$. Let’s take the map $$g : x \mapsto \cos \sqrt{x}$$. $$g$$ doesn’t have a limit at $$\infty$$ as for $$n \in \mathbb N$$, we have $$g(n^2 \pi^2) = \cos n \pi = (-1)^n$$. However, the derivative of $$g$$ is $g^\prime(x) = – \frac{\sin \sqrt{x}}{2 \sqrt{x}},$ and as $$\vert g^\prime(x) \vert \le \frac{1}{2 \sqrt{x}}$$ for all $$x \in (0,\infty)$$, we have $$\lim\limits_{x \to \infty} g^\prime(x) = 0$$.

# Limit points of real sequences

Let’s start by recalling an important theorem of real analysis:

THEOREM. A necessary and sufficient condition for the convergence of a real sequence is that it is bounded and has a unique limit point.

As a consequence of the theorem, a sequence having a unique limit point is divergent if it is unbounded. An example of such a sequence is the sequence $u_n = \frac{n}{2}(1+(-1)^n),$ whose initial values are $0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 6, \dots$ $$(u_n)$$ is an unbounded sequence whose unique limit point is $$0$$.

Let’s now look at sequences having more complicated limit points sets.

### A sequence whose set of limit points is the set of natural numbers

Consider the sequence $$(v_n)$$ whose initial terms are $1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, \dots$ $$(v_n)$$ is defined as follows $v_n=\begin{cases} 1 &\text{ for } n= 1\\ n – \frac{k(k+1)}{2} &\text{ for } \frac{k(k+1)}{2} \lt n \le \frac{(k+1)(k+2)}{2} \end{cases}$ $$(v_n)$$ is well defined as the sequence $$(\frac{k(k+1)}{2})_{k \in \mathbb N}$$ is strictly increasing with first term equal to $$1$$. $$(v_n)$$ is a sequence of natural numbers. As $$\mathbb N$$ is a set of isolated points of $$\mathbb R$$, we have $$V \subseteq \mathbb N$$, where $$V$$ is the set of limit points of $$(v_n)$$. Conversely, let’s take $$m \in \mathbb N$$. For $$k + 1 \ge m$$, we have $$\frac{k(k+1)}{2} + m \le \frac{(k+1)(k+2)}{2}$$, hence $u_{\frac{k(k+1)}{2} + m} = m$ which proves that $$m$$ is a limit point of $$(v_n)$$. Finally the set of limit points of $$(v_n)$$ is the set of natural numbers.

# Continuity versus uniform continuity

We consider real-valued functions.

A real-valued function $$f : I \to \mathbb R$$ (where $$I \subseteq$$ is an interval) is continuous at $$x_0 \in I$$ when: $(\forall \epsilon > 0) (\exists \delta > 0)(\forall x \in I)(\vert x- x_0 \vert \le \delta \Rightarrow \vert f(x)- f(x_0) \vert \le \epsilon).$ When $$f$$ is continuous at all $$x \in I$$, we say that $$f$$ is continuous on $$I$$.

$$f : I \to \mathbb R$$ is said to be uniform continuity on $$I$$ if $(\forall \epsilon > 0) (\exists \delta > 0)(\forall x,y \in I)(\vert x- y \vert \le \delta \Rightarrow \vert f(x)- f(y) \vert \le \epsilon).$

Obviously, a function which is uniform continuous on $$I$$ is continuous on $$I$$. Is the converse true? The answer is negative.

### An (unbounded) continuous function which is not uniform continuous

The map $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x^2 \end{array}$ is continuous. Let’s prove that it is not uniform continuous. For $$0 < x < y$$ we have $\vert f(x)-f(y) \vert = y^2-x^2 = (y-x)(y+x) \ge 2x (y-x)$ Hence for $$y-x= \delta >0$$ and $$x = \frac{1}{\delta}$$ we get
$\vert f(x) -f(y) \vert \ge 2x (y-x) =2 > 1$ which means that the definition of uniform continuity is not fulfilled for $$\epsilon = 1$$.

For this example, the function is unbounded as $$\lim\limits_{x \to \infty} x^2 = \infty$$. Continue reading Continuity versus uniform continuity

# No minimum at the origin but a minimum along all lines

We look here at an example, from the Italian mathematician Giuseppe Peano of a real function $$f$$ defined on $$\mathbb{R}^2$$. $$f$$ is having a local minimum at the origin along all lines passing through the origin, however $$f$$ does not have a local minimum at the origin as a function of two variables.

The function $$f$$ is defined as follows
$\begin{array}{l|rcl} f : & \mathbb{R}^2 & \longrightarrow & \mathbb{R} \\ & (x,y) & \longmapsto & f(x,y)=3x^4-4x^2y+y^2 \end{array}$ One can notice that $$f(x, y) = (y-3x^2)(y-x^2)$$. In particular, $$f$$ is strictly negative on the open set $$U=\{(x,y) \in \mathbb{R}^2 \ : \ x^2 < y < 3x^2\}$$, vanishes on the parabolas $$y=x^2$$ and $$y=3 x^2$$ and is strictly positive elsewhere. Consider a line $$D$$ passing through the origin. If $$D$$ is different from the coordinate axes, the equation of $$D$$ is $$y = \lambda x$$ with $$\lambda > 0$$. We have $f(x, \lambda x)= x^2(\lambda-3x)(\lambda -x).$ For $$x \in (-\infty,\frac{\lambda}{3}) \setminus \{0\}$$, $$f(x, \lambda x) > 0$$ while $$f(0,0)=0$$ which proves that $$f$$ has a local minimum at the origin along the line $$D \equiv y – \lambda x=0$$. Along the $$x$$-axis, we have $$f(x,0)=3 x^ 4$$ which has a minimum at the origin. And finally, $$f$$ also has a minimum at the origin along the $$y$$-axis as $$f(0,y)=y^2$$.

However, along the parabola $$\mathcal{P} \equiv y = 2 x^2$$ we have $$f(x,2 x^2)=-x^4$$ which is strictly negative for $$x \neq 0$$. As $$\mathcal{P}$$ is passing through the origin, $$f$$ assumes both positive and negative values in all neighborhood of the origin.

This proves that $$f$$ does not have a minimum at $$(0,0)$$.

# Counterexamples around Fubini’s theorem

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