# 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

# 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.

# Non commutative rings

Let’s recall that a set $$R$$ equipped with two operations $$(R,+,\cdot)$$ is a ring if and only if $$(R,+)$$ is an abelian group, multiplication $$\cdot$$ is associative and has a multiplicative identity $$1$$ and multiplication is left and right distributive with respect to addition.

$$(\mathbb Z, +, \cdot)$$ is a well known infinite ring which is commutative. The rational, real and complex numbers are other infinite commutative rings. Those are in fact fields as every non-zero element have a multiplicative inverse.

For a field $$F$$ (finite or infinite), the polynomial ring $$F[X]$$ is another example of infinite commutative ring.

Also for $$n$$ integer, the integers modulo n is a finite ring that is commutative. Finally, according to Wedderburn theorem every finite division ring is commutative.

So what are examples of non commutative rings? Let’s provide a couple. Continue reading Non commutative rings

# Around bounded sets, greatest element and supremum

Consider a linearly ordered set $$(X, \le)$$ and a subset $$S \subseteq X$$. Let’s recall some definitions:

• $$S$$ is bounded above if there exists an element $$k \in X$$ such that $$k \ge s$$ for all $$s \in S$$.
• $$g$$ is a greatest element of $$S$$ is $$g \in S$$ and $$g \ge s$$ for all $$s \in S$$. $$l$$ is a lowest element of $$S$$ is $$l \in S$$ and $$l \le s$$ for all $$s \in S$$.
• $$a$$ is an supremum of $$S$$ if it is the least element in $$X$$ that is greater than or equal to all elements of $$S$$.

### Subsets with a supremum but no greatest element

Let’s give examples of subsets having a supremum but no greatest element. First consider the ordered set $$(\mathbb R, \le)$$ and the subset $$S=\{ q \in \mathbb Q \ ; \ q \le \sqrt{2}\}$$. $$S$$ is bounded above by $$2$$. $$\sqrt{2}$$ is a supremum of $$S$$ as we have $$q \le \sqrt{2}$$ for all $$q \in S$$ and as for $$b < \sqrt{2}$$, it exists $$q \in \mathbb Q$$ such that $$b < q < \sqrt{2}$$ because $$\mathbb Q$$ is dense in $$\mathbb R$$. However $$S$$ doesn't have a greatest element because $$\sqrt{2}$$ is an irrational number.

For our second example we take $$X = \mathbb N \times \mathbb N$$ ordered lexicographically by $$\preceq$$. The subset $$S=\{(0,n) \ ; \ n \in \mathbb N\}$$ is bounded above by $$(2,0)$$. Moreover $$(1,0)$$ is a supremum. But $$S$$ doesn’t have a greatest element as for $$(0,n) \in S$$ we have $$(0,n) \prec (0,n+1)$$.

### Bounded above subsets with no supremum

Leveraging the examples above, we take $$(X, \le) = (\mathbb Q, \le)$$ and $$S=\{ q \in \mathbb Q \ ; \ q \le \sqrt{2}\}$$. $$S$$ is bounded above, by $$2$$ for example. However $$S$$ doesn’t have a supremum because $$\sqrt{2} \notin \mathbb Q$$.

Another example is the set $$X = \mathbb N \times \mathbb Z$$ ordered lexicographically by $$\preceq$$. The subset $$S=\{(0,n) \ ; \ n \in \mathbb N\}$$ is bounded above by $$(2,0)$$ but has no supremum. Indeed, the elements greater or equal to all the elements of $$S$$ are the elements $$(a,b)$$ with $$a \ge 1$$. However $$(a,b)$$ with $$a \ge 1$$ cannot be a supremum of $$S$$, as $$(a,b-1) \prec (a,b)$$ and $$(a,b-1)$$ is greater than all the elements of $$S$$.

# Counterexample around infinite products

Let’s recall two theorems about infinite products $$\prod \ (1+a_n)$$. The first one deals with nonnegative terms $$a_n$$.

THEOREM 1 An infinite product $$\prod \ (1+a_n)$$ with nonnegative terms $$a_n$$ converges if and only if the series $$\sum a_n$$ converges.

The second is related to infinite products with complex terms.

THEOREM 2 The absolute convergence of the series $$\sum a_n$$ implies the convergence of the infinite product $$\prod \ (1+a_n)$$. Moreover $$\prod \ (1+a_n)$$ is not zero providing $$a_n \neq -1$$ for all $$n \in \mathbb N$$.

The converse of Theorem 2 is not true as shown by following counterexample.

We consider $$a_n=(-1)^n/(n+1)$$. For $$N \in \mathbb N$$ we have:
$\prod_{n=1}^N \ (1+a_n) = \begin{cases} \frac{1}{2} &\text{ for } N \text{ odd}\\ \frac{1}{2}(1+\frac{1}{N+1}) &\text{ for } N \text{ even} \end{cases}$ hence the infinite product $$\prod \ (1+a_n)$$ converges (to $$\frac{1}{2}$$) while the series $$\sum \left\vert a_n \right\vert = \sum \frac{1}{n+1}$$ diverges (it is the harmonic series with first term omitted).