# A trigonometric series that is not a Fourier series (Riemann-integration)

We’re looking here at convergent trigonometric series like $f(x) = a_0 + \sum_{k=1}^\infty (a_n \cos nx + b_n \sin nx)$ which are convergent but are not Fourier series. Which means that the terms $$a_n$$ and $$b_n$$ cannot be written$\begin{array}{ll} a_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \cos nt \, dt & (n= 0, 1, \dots) \\ b_n = \frac{1}{\pi} \int_0^{2 \pi} g(t) \sin nt \, dt & (n= 1, 2, \dots) \end{array}$ where $$g$$ is any integrable function.

This raises the question of the type of integral used. We cover here an example based on Riemann integral. I’ll cover a Lebesgue integral example later on.

We prove here that the function $f(x)= \sum_{n=1}^\infty \frac{\sin nx}{\sqrt{n}}$ is a convergent trigonometric series but is not a Fourier series. Continue reading A trigonometric series that is not a Fourier series (Riemann-integration)

# 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

# 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

# Differentiability of multivariable real functions (part2)

Following the article on differentiability of multivariable real functions (part 1), we look here at second derivatives. We consider a function $$f : \mathbb R^n \to \mathbb R$$ with $$n \ge 2$$.

Schwarz’s theorem states that if $$f : \mathbb R^n \to \mathbb R$$ has continuous second partial derivatives at any given point in $$\mathbb R^n$$, then for $$(a_1, \dots, a_n) \in \mathbb R^n$$ and $$i,j \in \{1, \dots, n\}$$:
$\frac{\partial^2 f}{\partial x_i \partial x_j}(a_1, \dots, a_n)=\frac{\partial^2 f}{\partial x_j \partial x_i}(a_1, \dots, a_n)$

### A function for which $$\frac{\partial^2 f}{\partial x \partial y}(0,0) \neq \frac{\partial^2 f}{\partial y \partial x}(0,0)$$

We consider:
$\begin{array}{l|rcl} f : & \mathbb R^2 & \longrightarrow & \mathbb R \\ & (0,0) & \longmapsto & 0\\ & (x,y) & \longmapsto & \frac{xy(x^2-y^2)}{x^2+y^2} \text{ for } (x,y) \neq (0,0) \end{array}$ Continue reading Differentiability of multivariable real functions (part2)

# A discontinuous real convex function

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 $$1$$.

# Differentiability of multivariable real functions (part1)

This article provides counterexamples about differentiability of functions of several real variables. We focus on real functions of two real variables (defined on $$\mathbb R^2$$). $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. the absolute value for $$\mathbb R$$.

We recall some definitions and theorems about differentiability of functions of several real variables.

Definition 1 We say that a function $$f : \mathbb R^2 \to \mathbb R$$ is differentiable at $$\mathbf{a} \in \mathbb R^2$$ if it exists a (continuous) linear map $$\nabla f(\mathbf{a}) : \mathbb R^2 \to \mathbb R$$ with $\lim\limits_{\mathbf{h} \to 0} \frac{f(\mathbf{a}+\mathbf{h})-f(\mathbf{a})-\nabla f(\mathbf{a}).\mathbf{h}}{\Vert \mathbf{h} \Vert} = 0$

Definition 2 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. Then the $$\mathbf{i^{th}}$$ partial derivative at point $$\mathbf{a}$$ is the real number
\begin{align*}
\frac{\partial f}{\partial x_i}(\mathbf{a}) &= \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{e_i})- f(\mathbf{a})}{h}\\
&= \lim\limits_{h \to 0} \frac{f(a_1,\dots,a_{i-1},a_i+h,a_{i+1},\dots,a_n) – f(a_1,\dots,a_{i-1},a_i,a_{i+1},\dots,a_n)}{h}
\end{align*} For two real variable functions, $$\frac{\partial f}{\partial x}(x,y)$$ and $$\frac{\partial f}{\partial y}(x,y)$$ will denote the partial derivatives.

Definition 3 Let $$f : \mathbb R^n \to \mathbb R$$ be a real-valued function. The directional derivative of $$f$$ along vector $$\mathbf{v}$$ at point $$\mathbf{a}$$ is the real $\nabla_{\mathbf{v}}f(\mathbf{a}) = \lim\limits_{h \to 0} \frac{f(\mathbf{a}+h \mathbf{v})- f(\mathbf{a})}{h}$ Continue reading Differentiability of multivariable real functions (part1)

# Continuity of multivariable real functions

This article provides counterexamples about continuity of functions of several real variables. In addition the article discusses the cases of functions of two real variables (defined on $$\mathbb R^2$$ having real values. $$\mathbb R^2$$ and $$\mathbb R$$ are equipped with their respective Euclidean norms denoted by $$\Vert \cdot \Vert$$ and $$\vert \cdot \vert$$, i.e. the absolute value for $$\mathbb R$$.

We recall that a function $$f$$ defined from $$\mathbb R^2$$ to $$\mathbb R$$ is continuous at $$(x_0,y_0) \in \mathbb R^2$$ if for any $$\epsilon > 0$$, there exists $$\delta > 0$$, such that $$\Vert (x,y) -(x_0,y_0) \Vert < \delta \Rightarrow \vert f(x,y) - f(x_0,y_0) \vert < \epsilon$$. Continue reading Continuity of multivariable real functions

# Counterexamples on function limits (part 1)

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)

# Counterexamples around differentiation of sequences of functions

We consider here sequences of real functions defined on a closed interval. Following theorem is the main one regarding the differentiation of the limit.

Theorem: Suppose $$(f_n)$$ is a sequence of functions, differentiable on $$[a,b]$$ and such that $$(f_n(x_0))$$ converges for some point $$x_0 \in [a,b]$$. If $$(f_n^\prime)$$ converges uniformly on $$[a,b]$$, then $$(f_n)$$ converges uniformly on $$[a,b]$$ to a function $$f$$ and for all $$x \in [a,b]$$ $f^\prime(x)=\lim\limits_{n \to \infty} f_n^\prime(x)$ What happens if we drop some hypothesis of the theorem? Continue reading Counterexamples around differentiation of sequences of functions