Tag Archives: continuity

A monotonic function whose points of discontinuity form a dense set

Consider a compact interval \([a,b] \subset \mathbb R\) with \(a \lt b\). Let’s build an increasing function \(f : [a,b] \to \mathbb R\) whose points of discontinuity is an arbitrary dense subset \(D = \{d_n \ ; \ n \in \mathbb N\}\) of \([a,b]\), for example \(D = \mathbb Q \cap [a,b]\).

Let \(\sum p_n\) be a convergent series of positive numbers whose sum is equal to \(p\) and define \(\displaystyle f(x) = \sum_{d_n \le x} p_n\).

\(f\) is strictly increasing

For \(a \le x \lt y \le b\) we have \[
f(y) – f(x) = \sum_{x \lt d_n \le y} p_n \gt 0\] as the \(p_n\) are positive and dense so it exists \(p_m \in (x, y]\).

\(f\) is right-continuous on \([a,b]\)

We pick-up \(x \in [a,b]\). For any \(\epsilon \gt 0\) is exists \(N \in \mathbb N\) such that \(0 \lt \sum_{n \gt N} p_n \lt \epsilon\). Let \(\delta > 0\) be so small that the interval \((x,x+\delta)\) doesn’t contain any point in the finite set \(\{p_1, \dots, p_N\}\). Then \[
0 \lt f(y) – f(x) \le \sum_{n \gt N} p_n \lt \epsilon,\] for any \(y \in (x,x+\delta)\) proving the right-continuity of \(f\) at \(x\). Continue reading A monotonic function whose points of discontinuity form a dense set

Uniform continuous function but not Lipschitz continuous

Consider the function \[
\begin{array}{l|rcl}
f : & [0,1] & \longrightarrow & [0,1] \\
& x & \longmapsto & \sqrt{x} \end{array}\]

\(f\) is continuous on the compact interval \([0,1]\). Hence \(f\) is uniform continuous on that interval according to Heine-Cantor theorem. For a direct proof, one can verify that for \(\epsilon > 0\), one have \(\vert \sqrt{x} – \sqrt{y} \vert \le \epsilon\) for \(\vert x – y \vert \le \epsilon^2\).

However \(f\) is not Lipschitz continuous. If \(f\) was Lipschitz continuous for a Lipschitz constant \(K > 0\), we would have \(\vert \sqrt{x} – \sqrt{y} \vert \le K \vert x – y \vert\) for all \(x,y \in [0,1]\). But we get a contradiction taking \(x=0\) and \(y=\frac{1}{4 K^2}\) as \[
\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert\]

Pointwise convergence not uniform on any interval

We provide in this article an example of a pointwise convergent sequence of real functions that doesn’t converge uniformly on any interval.

Let’s consider a sequence \((a_p)_{p \in \mathbb N}\) enumerating the set \(\mathbb Q\) of rational numbers. Such a sequence exists as \(\mathbb Q\) is countable.

Now let \((g_n)_{n \in \mathbb N}\) be the sequence of real functions defined on \(\mathbb R\) by \[
g_n(x) = \sum_{p=1}^{\infty} \frac{1}{2^p} f_n(x-a_p)\] where \(f_n : x \mapsto \frac{n^2 x^2}{1+n^4 x^4}\) for \(n \in \mathbb N\).

\(f_n\) main properties

\(f_n\) is a rational function whose denominator doesn’t vanish. Hence \(f_n\) is indefinitely differentiable. As \(f_n\) is an even function, we can study it only on \([0,\infty)\).

We have \[
f_n^\prime(x)= 2n^2x \frac{1-n^4x^4}{(1+n^4 x^4)^2}.\] \(f_n^\prime\) vanishes at zero (like \(f_n\)) is positive on \((0,\frac{1}{n})\), vanishes at \(\frac{1}{n}\) and is negative on \((\frac{1}{n},\infty)\). Hence \(f_n\) has a maximum at \(\frac{1}{n}\) with \(f_n(\frac{1}{n}) = \frac{1}{2}\) and \(0 \le f_n(x) \le \frac{1}{2}\) for all \(x \in \mathbb R\).

Also for \(x \neq 0\) \[
0 \le f_n(x) =\frac{n^2 x^2}{1+n^4 x^4} \le \frac{n^2 x^2}{n^4 x^4} = \frac{1}{n^2 x^2}\] consequently \[
0 \le f_n(x) \le \frac{1}{n} \text{ for } x \ge \frac{1}{\sqrt{n}}.\]

\((g_n)\) converges pointwise to zero

First, one can notice that \(g_n\) is well defined. For \(x \in \mathbb R\) and \(p \in \mathbb N\) we have \(0 \le \frac{1}{2^p} f_n(x-a_p) \le \frac{1}{2^p} \cdot\ \frac{1}{2}=\frac{1}{2^{p+1}}\) according to previous paragraph. Therefore the series of functions \(\sum \frac{1}{2^p} f_n(x-a_p)\) is normally convergent. \(g_n\) is also continuous as for all \(p \in \mathbb N\) \(x \mapsto \frac{1}{2^p} f_n(x-a_p)\) is continuous. Continue reading Pointwise convergence not uniform on any interval

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 discontinuous additive map

A function \(f\) defined on \(\mathbb R\) into \(\mathbb R\) is said to be additive if and only if for all \(x, y \in \mathbb R\)
\[f(x+y) = f(x) + f(y).\] If \(f\) is supposed to be continuous at zero, \(f\) must have the form \(f(x)=cx\) where \(c=f(1)\). This can be shown using following steps:

  • \(f(0) = 0\) as \(f(0) = f(0+0)= f(0)+f(0)\).
  • For \(q \in \mathbb N\) \(f(1)=f(q \cdot \frac{1}{q})=q f(\frac{1}{q})\). Hence \(f(\frac{1}{q}) = \frac{f(1)}{q}\). Then for \(p,q \in \mathbb N\), \(f(\frac{p}{q}) = p f(\frac{1}{q})= f(1) \frac{p}{q}\).
  • As \(f(-x) = -f(x)\) for all \(x \in\mathbb R\), we get that for all rational number \(\frac{p}{q} \in \mathbb Q\), \(f(\frac{p}{q})=f(1)\frac{p}{q}\).
  • The equality \(f(x+y) = f(x) + f(y)\) implies that \(f\) is continuous on \(\mathbb R\) if it is continuous at \(0\).
  • We can finally conclude to \(f(x)=cx\) for all real \(x \in \mathbb R\) as the rational numbers are dense in \(\mathbb R\).

We’ll use a Hamel basis to construct a discontinuous linear function. The set \(\mathbb R\) can be endowed with a vector space structure over \(\mathbb Q\) using the standard addition and the multiplication by a rational for the scalar multiplication.

Using the axiom of choice, one can find a (Hamel) basis \(\mathcal B = (b_i)_{i \in I}\) of \(\mathbb R\) over \(\mathbb Q\). That means that every real number \(x\) is a unique linear combination of elements of \(\mathcal B\): \[
x= q_1 b_{i_1} + \dots + q_n b_{i_n}\] with rational coefficients \(q_1, \dots, q_n\). The function \(f\) is then defined as \[
f(x) = q_1 + \dots + q_n.\] The linearity of \(f\) follows from its definition. \(f\) is not continuous as it only takes rational values which are not all equal. And one knows that the image of \(\mathbb R\) under a continuous map is an interval.

Existence of a continuous function with divergent Fourier series

In that article, I provided an example of a continuous function with divergent Fourier series. We prove here the existence of such a function using Banach-Steinhaus theorem, also called uniform boundedness principle.

Theorem (Uniform Boundedness Theorem) Let \((X, \Vert \cdot \Vert_X)\) be a Banach space and \((Y, \Vert \cdot \Vert_Y)\) be a normed vector space. Suppose that \(F\) is a set of continuous linear operators from \(X\) to \(Y\). If for all \(x \in X\) one has \[
\sup\limits_{T \in F} \Vert T(x) \Vert_Y \lt \infty\] then \[
\sup\limits_{T \in F, \ \Vert x \Vert = 1} \Vert T(x) \Vert_Y \lt \infty\]

Let’s take for \(X\) the vector space \(\mathcal C_{2 \pi}\) of continuous functions from \(\mathbb R\) to \(\mathbb C\) which are periodic with period \(2 \pi\) endowed with the norm \(\Vert f \Vert_\infty = \sup\limits_{- \pi \le t \le \pi} \vert f(t) \vert\). \((\mathcal C_{2 \pi}, \Vert \cdot \Vert_\infty)\) is a Banach space. For the vector space \(Y\), we take the complex numbers \(\mathbb C\) endowed with the modulus.

For \(n \in \mathbb N\), the map \[
\begin{array}{l|rcl}
\ell_n : & \mathcal C_{2 \pi} & \longrightarrow & \mathbb C \\
& f & \longmapsto & \displaystyle \sum_{p=-n}^n c_p(f) \end{array}\] is a linear operator, where for \(p \in \mathbb Z\), \(c_p(f)\) denotes the complex Fourier coefficient \[
c_p(f) = \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(t) e^{-i p t} \ dt\]

We now prove that
\begin{align*}
\Lambda_n &= \sup\limits_{f \in \mathcal C_{2 \pi}, \Vert f \Vert_\infty=1} \vert \ell_n(f) \vert\\
&= \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} \right\vert \ dt = \frac{1}{2 \pi} \int_{- \pi}^{\pi} \left\vert h_n(t) \right\vert \ dt,
\end{align*} where one can notice that the function \[
\begin{array}{l|rcll}
h_n : & [- \pi, \pi] & \longrightarrow & \mathbb C \\
& t & \longmapsto & \frac{\sin (2n+1)\frac{t}{2}}{\sin \frac{t}{2}} &\text{for } t \neq 0\\
& 0 & \longmapsto & 2n+1
\end{array}\] is continuous.
Continue reading Existence of a continuous function with divergent Fourier series

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 continuous function with divergent Fourier series

It is known that for a piecewise continuously differentiable function \(f\), the Fourier series of \(f\) converges at all \(x \in \mathbb R\) to \(\frac{f(x^-)+f(x^+)}{2}\).

We describe Fejér example of a continuous function with divergent Fourier series. Fejér example is the even, \((2 \pi)\)-periodic function \(f\) defined on \([0,\pi]\) by: \[
f(x) = \sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]\]
According to Weierstrass M-test, \(f\) is continuous. We denote \(f\) Fourier series by \[
\frac{1}{2} a_0 + (a_1 \cos x + b_1 \sin x) + \dots + (a_n \cos nx + b_n \sin nx) + \dots.\]

As \(f\) is even, the \(b_n\) are all vanishing. If we denote for all \(m \in \mathbb N\):\[
\lambda_{n,m}=\int_0^{\pi} \sin \left[ (2m + 1) \frac{t}{2} \right] \cos nt \ dt \text{ and } \sigma_{n,m} = \sum_{k=0}^n \lambda_{k,m},\]
we have:\[
\begin{aligned}
a_n &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos nt \ dt= \frac{2}{\pi} \int_0^{\pi} f(t) \cos nt \ dt\\
&= \frac{2}{\pi} \int_0^{\pi} \left(\sum_{p=1}^\infty \frac{1}{p^2} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right]\right) \cos nt \ dt\\
&=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \int_0^{\pi} \sin \left[ (2^{p^3} + 1) \frac{x}{2} \right] \cos nt \ dt\\
&=\frac{2}{\pi} \sum_{p=1}^\infty \frac{1}{p^2} \lambda_{n,2^{p^3-1}}
\end{aligned}\] One can switch the \(\int\) and \(\sum\) signs as the series is normally convergent.

We now introduce for all \(n \in \mathbb N\):\[
S_n = \frac{\pi}{2} \sum_{k=0}^n a_k = \sum_{p=1}^\infty \sum_{k=0}^n \frac{1}{p^2} \lambda_{k,2^{p^3-1}}
=\sum_{p=1}^\infty \frac{1}{p^2} \sigma_{n,2^{p^3-1}}\]

We will prove below that for all \(n,m \in \mathbb N\) we have \(\sigma_{m,m} \ge \frac{1}{2} \ln m\) and \(\sigma_{n,m} \ge 0\). Assuming those inequalities for now, we get:\[
S_{2^{p^3-1}} \ge \frac{1}{p^2} \sigma_{2^{p^3-1},2^{p^3-1}} \ge \frac{1}{2p^2} \ln(2^{p^3-1}) = \frac{p^3-1}{2p^2} \ln 2\]
As the right hand side diverges to \(\infty\), we can conclude that \((S_n)\) diverges and consequently that the Fourier series of \(f\) diverges at \(0\). Continue reading A continuous function with divergent Fourier series

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

Continuity under the integral sign

We consider here a measure space \((\Omega, \mathcal A, \mu)\) and \(T \subset \mathbb R\) a topological subspace. For a map \(f : T \times \Omega \to \mathbb R\) such that for all \(t \in T\) the map \[
\begin{array}{l|rcl}
f(t, \cdot) : & \Omega & \longrightarrow & \mathbb R \\
& \omega & \longmapsto & f(t,\omega) \end{array}
\] is integrable, one can define the function \[
\begin{array}{l|rcl}
F : & T & \longrightarrow & \mathbb R \\
& t & \longmapsto & \int_\Omega f(t,\omega) \ d\mu(\omega) \end{array}
\]

Following theorem is well known (and can be proven using dominated convergence theorem):

THEOREM for an adherent point \(x \in T\), if

  • \(\forall \omega \in \Omega \lim\limits_{t \to x} f(t,\omega) = \varphi(\omega)\)
  • There exists a map \(g : \Omega \to \mathbb R\) such that \(\forall t \in T, \, \forall \omega \in \Omega, \ \vert f(t,\omega) \vert \le g(\omega)\)

then \(\varphi\) is integrable and \[
\lim\limits_{t \to x} F(t) = \int_\Omega \varphi(\omega) \ d\mu(\omega)\]
In other words, one can switch \(\lim\) and \(\int\) signs.

We provide here a counterexample showing that the conclusion of the theorem might not hold if \(f\) is not bounded by a function \(g\) as supposed in the premises of the theorem. Continue reading Continuity under the integral sign