Tag Archives: derivative

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 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 \(\lim\limits_{x \to \infty} g^\prime(x) = 0\).

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

A function whose derivative at 0 is one but which is not increasing near 0

From the mean value theorem, a real function whose derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function \(f\) defined in a non-degenerate interval \(I\) with a strictly positive derivative at a point \(a\) of the interval is strictly increasing near that point. For the proof, we just have to notice that as \(f^\prime\) is continuous and \(f^\prime(a) > 0\), \(f^\prime\) is strictly positive within an interval \(J \subset I\) containing \(a\). By the mean value theorem, \(f\) is strictly increasing on \(J\).

We now suppose that \(f\) is differentiable on an interval \(I\) containing \(0\) with \(f^\prime(0)>0\). For \(x>0\) sufficiently close to zero we have \(\displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0\), hence \(f(x)>f(0)\). But that doesn’t imply that \(f\) is strictly increasing in a neighborhood of zero. Let’s prove it with a counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0

A decreasing function converging to zero whose derivative diverges (part2)

In that article, I gave examples of real valued functions defined on \((0,+\infty)\) that converge to zero and whose derivatives diverge. But those functions were not monotonic. Here I give an example of a decreasing real valued function \(g\) converging to zero at \(+\infty\) and whose derivative is unbounded.

We first consider the polynomial map:
\[P(x)=(1+2x)(1-x)^2=1-3x^2+2x^3\] on the segment \(I=[0,1]\). \(P\) derivative equals \(P^\prime(x)=-6x(1-x)\). Therefore \(P\) is decreasing on \(I\). Moreover we have \(P(0)=1\), \(P(1)=P^\prime(0)=P^\prime(1)=0\) and \(P^\prime(1/2)=-3/2\). Continue reading A decreasing function converging to zero whose derivative diverges (part2)

Differentiable functions converging to zero whose derivatives diverge (part1)

In this article, I consider real valued functions \(f\) defined on \((0,+\infty)\) that converge to zero, i.e.:
\[\lim\limits_{x \to +\infty} f(x) = 0\] If \(f\) is differentiable what can be the behavior of its derivative as \(x\) approaches \(+\infty\)?

Let’s consider a first example:
\[\begin{array}{l|rcl}
f_1 : & (0,+\infty) & \longrightarrow & \mathbb{R} \\
& x & \longmapsto & \frac{1}{x} \end{array}\] \(f_1\) derivative is \(f_1^\prime(x)=-\frac{1}{x^2}\) and we also have \(\lim\limits_{x \to +\infty} f_1^\prime(x) = 0\). Let’s consider more sophisticated cases! Continue reading Differentiable functions converging to zero whose derivatives diverge (part1)

A continuous function not differentiable at the rationals but differentiable elsewhere

We build here a continuous function of one real variable whose derivative exists on \(\mathbb{R} \setminus \mathbb{Q}\) and doesn’t have a left or right derivative on each point of \(\mathbb{Q}\).

As \(\mathbb{Q}\) is (infinitely) countable, we can find a bijection \(n \mapsto r_n\) from \(\mathbb{N}\) to \(\mathbb{Q}\). We now reuse the function \(f\) defined here. Recall \(f\) main properties: Continue reading A continuous function not differentiable at the rationals but differentiable elsewhere

A differentiable function except at one point with a bounded derivative

We build here a continuous function of one real variable whose derivative exists except at \(0\) and is bounded on \(\mathbb{R^*}\).

We start with the even and piecewise linear function \(g\) defined on \([0,+\infty)\) with following values:
\[g(x)=
\left\{
\begin{array}{ll}
0 & \mbox{if } x =0\\
0 & \mbox{if } x \in \{\frac{k}{4^n};(k,n) \in \{1,2,4\} \times \mathbb{N^*}\}\\
1 & \mbox{if } x \in \{\frac{3}{4^n};n \in \mathbb{N^*}\}\\
\end{array}
\right.
\]
The picture below gives an idea of the graph of \(g\) for positive values. Continue reading A differentiable function except at one point with a bounded derivative