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 \(1\).
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→
We look here at the continuity of a sequence of functions that converges pointwise and give some counterexamples of what happens versus uniform convergence.
Recalling the definition of pointwise convergence
We consider here real functions defined on a closed interval \([a,b]\). A sequence of functions \((f_n)\) defined on \([a,b]\) converges pointwise to the function \(f\) if and only if for all \(x \in [a,b]\) \(\displaystyle \lim\limits_{n \to +\infty} f_n(x) = f(x)\). Pointwise convergence is weaker than uniform convergence.
Pointwise convergence does not, in general, preserve continuity
Suppose that \(f_n \ : \ [0,1] \to \mathbb{R}\) is defined by \(f_n(x)=x^n\). For \(0 \le x <1\) then \(\displaystyle \lim\limits_{n \to +\infty} x^n = 0\), while if \(x = 1\) then \(\displaystyle \lim\limits_{n \to +\infty} x^n = 1\). Hence the sequence \(f_n\) converges to the function equal to \(0\) for \(0 \le x < 1\) and to \(1\) for \(x=1\).
Although each \(f_n\) is a continuous function of \([0,1]\), their pointwise limit is not. \(f\) is discontinuous at \(1\).
We notice that \((f_n)\) doesn't converge uniformly to \(f\) as for all \(n \in \mathbb{N}\), \(\displaystyle \sup\limits_{x \in [0,1]} \vert f_n(x) - f(x) \vert = 1\). That's reassuring as uniform convergence of a sequence of continuous functions implies that the limit is continuous! Continue reading Pointwise convergence and properties of the limit (part 1)→
In that article, I described some properties of Thomae’s function\(f\). Namely:
The function is discontinuous on \(\mathbb{Q}\).
Continuous on \(\mathbb{R} \setminus \mathbb{Q}\).
Its right-sided and left-sided limits vanish at all points.
Let’s modify \(f\) to get function \(g\) defined as follow:
\[g:
\left|\begin{array}{lrl}
\mathbb{R} & \longrightarrow & \mathbb{R} \\
x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\
\frac{p}{q} & \longmapsto & q \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0
\end{array}\right.\] \(f\) and \(g\) both vanish on the set of irrational numbers, while on the set of rational numbers, \(g\) is equal to the reciprocal of \(f\). We now consider an open subset \(O \subset \mathbb{R}\) and \(x \in O\). As \(f\) right-sided and left-sided limits vanish at all points, we have \(\lim\limits_{n \to +\infty} f(x_n) = 0\) for all sequence \((x_n)\) of rational numbers converging to \(x\) (and such a sequence exists as the rational numbers are everywhere dense in the reals). Hence \(\lim\limits_{n \to +\infty} g(x_n) = + \infty\) as \(f\) is positive.
We can conclude that \(g\) is nowhere locally bounded. The picture of the article is a plot of function \(g\) on the rational numbers \(r = \frac{p}{q}\) in lowest terms for \(0 < r < 1\) and \(q \le 50\).
Let’s discover the beauties of Thomae’s function also named the popcorn function, the raindrop function or the modified Dirichlet function.
Thomae’s function is a real-valued function defined as:
\[f:
\left|\begin{array}{lrl}
\mathbb{R} & \longrightarrow & \mathbb{R} \\
x & \longmapsto & 0 \text{ if } x \in \mathbb{R} \setminus \mathbb{Q} \\
\frac{p}{q} & \longmapsto & \frac{1}{q} \text{ if } \frac{p}{q} \text{ in lowest terms and } q > 0
\end{array}\right.\]
\(f\) is periodic with period \(1\)
This is easy to prove as for \(x \in \mathbb{R} \setminus \mathbb{Q}\) we also have \(x+1 \in \mathbb{R} \setminus \mathbb{Q}\) and therefore \(f(x+1)=f(x)=0\). While for \(y=\frac{p}{q} \in \mathbb{Q}\) in lowest terms, \(y+1=\frac{p+q}{q}\) is also in lowest terms, hence \(f(y+1)=f(y)=\frac{1}{q}\). Continue reading A function continuous at all irrationals and discontinuous at all rationals→
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)→
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)→
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 rightderivative on each point of \(\mathbb{Q}\).
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→
Recall that a function of bounded variation, also known as a BV-function, is a real-valued function whose total variation is bounded (finite).
Being more formal, the total variation of a real-valued function \(f\), defined on an interval \([a,b] \subset \mathbb{R}\) is the quantity:
\[V_a^b(f) = \sup\limits_{P \in \mathcal{P}} \sum_{i=0}^{n_P-1} \left\vert f(x_{i+1}) – f(x_i) \right\vert\] where the supremum is taken over the set \(\mathcal{P}\) of all partitions of the interval considered. Continue reading A continuous function which is not of bounded variation→
