# A nonzero continuous map orthogonal to all polynomials

Let’s consider the vector space $$\mathcal{C}^0([a,b],\mathbb R)$$ of continuous real functions defined on a compact interval $$[a,b]$$. We can define an inner product on pairs of elements $$f,g$$ of $$\mathcal{C}^0([a,b],\mathbb R)$$ by $\langle f,g \rangle = \int_a^b f(x) g(x) \ dx.$

It is known that $$f \in \mathcal{C}^0([a,b],\mathbb R)$$ is the always vanishing function if we have $$\langle x^n,f \rangle = \int_a^b x^n f(x) \ dx = 0$$ for all integers $$n \ge 0$$. Let’s recall the proof. According to Stone-Weierstrass theorem, for all $$\epsilon >0$$ if exists a polynomial $$P$$ such that $$\Vert f – P \Vert_\infty \le \epsilon$$. Then \begin{aligned} 0 &\le \int_a^b f^2 = \int_a^b f(f-P) + \int_a^b fP\\ &= \int_a^b f(f-P) \le \Vert f \Vert_\infty \epsilon(b-a) \end{aligned} As this is true for all $$\epsilon > 0$$, we get $$\int_a^b f^2 = 0$$ and $$f = 0$$.

We now prove that the result becomes false if we change the interval $$[a,b]$$ into $$[0, \infty)$$, i.e. that one can find a continuous function $$f \in \mathcal{C}^0([0,\infty),\mathbb R)$$ such that $$\int_0^\infty x^n f(x) \ dx$$ for all integers $$n \ge 0$$. In that direction, let’s consider the complex integral $I_n = \int_0^\infty x^n e^{-(1-i)x} \ dx.$ $$I_n$$ is well defined as for $$x \in [0,\infty)$$ we have $$\vert x^n e^{-(1-i)x} \vert = x^n e^{-x}$$ and $$\int_0^\infty x^n e^{-x} \ dx$$ converges. By integration by parts, one can prove that $I_n = \frac{n!}{(1-i)^{n+1}} = \frac{(1+i)^{n+1}}{2^{n+1}} n! = \frac{e^{i \frac{\pi}{4}(n+1)}}{2^{\frac{n+1}{2}}}n!.$ Consequently, $$I_{4p+3} \in \mathbb R$$ for all $$p \ge 0$$ which means $\int_0^\infty x^{4p+3} \sin(x) e^{-x} \ dx =0$ and finally $\int_0^\infty u^p \sin(u^{1/4}) e^{-u^{1/4}} \ dx =0$ for all integers $$p \ge 0$$ using integration by substitution with $$x = u^{1/4}$$. The function $$u \mapsto \sin(u^{1/4}) e^{-u^{1/4}}$$ is one we were looking for.

# Counterexamples on real sequences (part 3)

Let $$(u_n)$$ be a sequence of real numbers.

### If $$u_{2n}-u_n \le \frac{1}{n}$$ then $$(u_n)$$ converges?

This is wrong. The sequence
$u_n=\begin{cases} 0 & \text{for } n \notin \{2^k \ ; \ k \in \mathbb N\}\\ 1- 2^{-k} & \text{for } n= 2^k\end{cases}$
is a counterexample. For $$n \gt 2$$ and $$n \notin \{2^k \ ; \ k \in \mathbb N\}$$ we also have $$2n \notin \{2^k \ ; \ k \in \mathbb N\}$$, hence $$u_{2n}-u_n=0$$. For $$n = 2^k$$ $0 \le u_{2^{k+1}}-u_{2^k}=2^{-k}-2^{-k-1} \le 2^{-k} = \frac{1}{n}$ and $$\lim\limits_{k \to \infty} u_{2^k} = 1$$. $$(u_n)$$ does not converge as $$0$$ and $$1$$ are limit points.

### If $$\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$$ then $$(u_n)$$ has a finite or infinite limit?

This is not true. Let’s consider the sequence
$u_n=2+\sin(\ln n)$ Using the inequality $$\vert \sin p – \sin q \vert \le \vert p – q \vert$$
which is a consequence of the mean value theorem, we get $\vert u_{n+1} – u_n \vert = \vert \sin(\ln (n+1)) – \sin(\ln n) \vert \le \vert \ln(n+1) – \ln(n) \vert$ Therefore $$\lim\limits_n \left(u_{n+1}-u_n \right) =0$$ as $$\lim\limits_n \left(\ln(n+1) – \ln(n)\right) = 0$$. And $$\lim\limits_{n} \frac{u_{n+1}}{u_n} =1$$ because $$u_n \ge 1$$ for all $$n \in \mathbb N$$.

I now assert that the interval $$[1,3]$$ is the set of limit points of $$(u_n)$$. For the proof, it is sufficient to prove that $$[-1,1]$$ is the set of limit points of the sequence $$v_n=\sin(\ln n)$$. For $$y \in [-1,1]$$, we can pickup $$x \in \mathbb R$$ such that $$\sin x =y$$. Let $$\epsilon > 0$$ and $$M \in \mathbb N$$ , we can find an integer $$N \ge M$$ such that $$0 < \ln(n+1) - \ln(n) \lt \epsilon$$ for $$n \ge N$$. Select $$k \in \mathbb N$$ with $$x +2k\pi \gt \ln N$$ and $$N_\epsilon$$ with $$\ln N_\epsilon \in (x +2k\pi, x +2k\pi + \epsilon)$$. This is possible as $$(\ln n)_{n \in \mathbb N}$$ is an increasing sequence and the length of the interval $$(x +2k\pi, x +2k\pi + \epsilon)$$ is equal to $$\epsilon$$. We finally get $\vert u_{N_\epsilon} - y \vert = \vert \sin \left(\ln N_\epsilon \right) - \sin \left(x + 2k \pi \right) \vert \le \left(\ln N_\epsilon - (x +2k\pi)\right) \le \epsilon$ proving that $$y$$ is a limit point of $$(u_n)$$.

# A strictly increasing continuous function that is differentiable at no point of a null set

We build in this article a strictly increasing continuous function $$f$$ that is differentiable at no point of a null set $$E$$. The null set $$E$$ can be chosen arbitrarily. In particular it can have the cardinality of the continuum like the Cantor null set.

### A set of strictly increasing continuous functions

For $$p \lt q$$ two real numbers, consider the function $f_{p,q}(x)=(q-p) \left[\frac{\pi}{2} + \arctan{\left(\frac{2x-p-q}{q-p}\right)}\right]$ $$f_{p,q}$$ is positive and its derivative is $f_{p,q}^\prime(x) = \frac{2}{1+\left(\frac{2x-p-q}{q-p}\right)^2}$ which is always strictly positive. Hence $$f_{p,q}$$ is strictly increasing. We also have $\lim\limits_{x \to -\infty} f_{p,q}(x) = 0 \text{ and } \lim\limits_{x \to \infty} f_{p,q}(x) = \pi(q-p).$ One can notice that for $$x \in (p,q)$$, $$f_{p,q}^\prime(x) \gt 1$$. Therefore for $$x, y \in (p,q)$$ distinct we have according to the mean value theorem $$\frac{f_{p,q}(y)-f_{p,q}(x)}{y-x} \ge 1$$.

### Covering $$E$$ with an appropriate set of open intervals

As $$E$$ is a null set, for each $$n \in \mathbb N$$ one can find an open set $$O_n$$ containing $$E$$ and measuring less than $$2^{-n}$$. $$O_n$$ can be written as a countable union of disjoint open intervals as any open subset of the reals. Then $$I=\bigcup_{m \in \mathbb N} O_m$$ is also a countable union of open intervals $$I_n$$ with $$n \in \mathbb N$$. The sum of the lengths of the $$I_n$$ is less than $$1$$. Continue reading A strictly increasing continuous function that is differentiable at no point of a null set

# A function whose Maclaurin series converges only at zero

Let’s describe a real function $$f$$ whose Maclaurin series converges only at zero. For $$n \ge 0$$ we denote $$f_n(x)= e^{-n} \cos n^2x$$ and $f(x) = \sum_{n=0}^\infty f_n(x)=\sum_{n=0}^\infty e^{-n} \cos n^2 x.$ For $$k \ge 0$$, the $$k$$th-derivative of $$f_n$$ is $f_n^{(k)}(x) = e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right)$ and $\left\vert f_n^{(k)}(x) \right\vert \le e^{-n} n^{2k}$ for all $$x \in \mathbb R$$. Therefore $$\displaystyle \sum_{n=0}^\infty f_n^{(k)}(x)$$ is normally convergent and $$f$$ is an indefinitely differentiable function with $f^{(k)}(x) = \sum_{n=0}^\infty e^{-n} n^{2k} \cos \left(n^2 x + \frac{k \pi}{2}\right).$ Its Maclaurin series has only terms of even degree and the absolute value of the term of degree $$2k$$ is $\left(\sum_{n=0}^\infty e^{-n} n^{4k}\right)\frac{x^{2k}}{(2k)!} > e^{-2k} (2k)^{4k}\frac{x^{2k}}{(2k)!} > \left(\frac{2kx}{e}\right)^{2k}.$ The right hand side of this inequality is greater than $$1$$ for $$k \ge \frac{e}{2x}$$. This means that for any nonzero $$x$$ the Maclaurin series for $$f$$ diverges.

# 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 $\vert \sqrt{x} – \sqrt{y} \vert = \frac{1}{2 K} > \frac{1}{4 K} = K \vert x – y \vert$

# Counterexamples around Cauchy condensation test

According to Cauchy condensation test: for a non-negative, non-increasing sequence $$(u_n)_{n \in \mathbb N}$$ of real numbers, the series $$\sum_{n \in \mathbb N} u_n$$ converges if and only if the condensed series $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ converges.

The test doesn’t hold for any non-negative sequence. Let’s have a look at counterexamples.

### A sequence such that $$\sum_{n \in \mathbb N} u_n$$ converges and $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges

Consider the sequence $u_n=\begin{cases} \frac{1}{n} & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ 0 & \text{ else} \end{cases}$ For $$n \in \mathbb N$$ we have $0 \le \sum_{k = 1}^n u_k \le \sum_{k = 1}^{2^n} u_k = \sum_{k = 1}^{n} \frac{1}{2^k} < 1,$ therefore $$\sum_{n \in \mathbb N} u_n$$ converges as its partial sums are positive and bounded above. However $\sum_{k=1}^n 2^k u_{2^k} = \sum_{k=1}^n 1 = n,$ so $$\sum_{n \in \mathbb N} 2^n u_{2^n}$$ diverges.

### A sequence such that $$\sum_{n \in \mathbb N} v_n$$ diverges and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges

Consider the sequence $v_n=\begin{cases} 0 & \text{ for } n \in \{2^k \ ; \ k \in \mathbb N\}\\ \frac{1}{n} & \text{ else} \end{cases}$ We have $\sum_{k = 1}^{2^n} v_k = \sum_{k = 1}^{2^n} \frac{1}{k} – \sum_{k = 1}^{n} \frac{1}{2^k} > \sum_{k = 1}^{2^n} \frac{1}{k} -1$ which proves that the series $$\sum_{n \in \mathbb N} v_n$$ diverges as the harmonic series is divergent. However for $$n \in \mathbb N$$, $$2^n v_{2^n} = 0$$ and $$\sum_{n \in \mathbb N} 2^n v_{2^n}$$ converges.

# 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 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 $\lim\limits_{n \to \infty} \vert f(x_n) \vert = \lim\limits_{n \to \infty} \vert f(q_n^1 b_{i_1} + q_n^2 b_{i_2}) \vert = \lim\limits_{n \to \infty} \vert q_n^1 \vert = \infty.$