Tag Archives: integration

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 \[
0 &\le \int_a^b f^2 = \int_a^b f(f-P) + \int_a^b fP\\
&= \int_a^b f(f-P) \le \Vert f \Vert_\infty \epsilon(b-a)
\end{aligned}\] 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.

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.

Counterexamples around Lebesgue’s Dominated Convergence Theorem

Let’s recall Lebesgue’s Dominated Convergence Theorem. Let \((f_n)\) be a sequence of real-valued measurable functions on a measure space \((X, \Sigma, \mu)\). Suppose that the sequence converges pointwise to a function \(f\) and is dominated by some integrable function \(g\) in the sense that \[
\vert f_n(x) \vert \le g (x)\] for all \(n \in \mathbb N\) and all \(x \in X\).
Then \(f\) is integrable and \[
\lim\limits_{n \to \infty} \int_X f_n(x) \ d \mu = \int_X f(x) \ d \mu\]

Let’s see what can happen if we drop the domination condition.

We consider the space \(\mathbb R\) endowed with Lebesgue measure and for \(E \subseteq \mathbb R\) we denote by \(\chi_E\) the indicator function of \(E\) defined by \[
1 \text{ if } x \in E\\
0 \text{ otherwise}\end{cases}\] For \(n \in \mathbb N\), the function \(f_n=\frac{1}{2n}\chi_{(n^2-n,n^2+n)}\) is measurable and we have \[
\int_{\mathbb R} \frac{1}{2n}\chi_{(n^2-n,n^2+n)}(x) \ dx = \int_{n^2-n}^{n^2+n} \frac{1}{2n} \ dx = 1\] The sequence \((f_n)\) converges uniformly (and therefore pointwise) to the always vanishing function as for \(n \in \mathbb N\) we have for all \(x \in \mathbb R\) \(\vert f_n(x) \vert \le \frac{1}{2n}\). Hence the conclusion of Lebesgue’s Dominated Convergence Theorem doesn’t hold for the sequence \((f_n)\).

Let’s verify that the sequence \((f_n)\) is not dominated by some integrable function \(g\). For \(p < q\) integers, we have \[ \begin{aligned} q^2-q-(p^2+p) &= q^2-p^2 -q-p\\ &= (q-p)(q+p) -q -p\\ &\ge (q+p) -q-p=0 \end{aligned}\] Hence for \(p \neq q\) integers the intervals \((p^2-p,p^2+p)\) and \((q^2-q,q^2+q)\) are disjoint. Consequently for all \(x \in \mathbb R\) the sum \(\sum_{n \in \mathbb N} f_n(x)\) amounts to only one term and the function \(\sum_{n \in \mathbb N} f_n\) is well defined. If \(g\) dominates the sequence \((f_n)\), it satisfies \(0 \le \sum_{n \in \mathbb N} f_n \le g\). But \[ \int_{\mathbb R} \sum_{n \in \mathbb N} f_n(x) \ dx = \sum_{n \in \mathbb N} \int_{\mathbb R} f_n(x) \ dx = \sum_{n \in \mathbb N} 1 = \infty\] and \(g\) cannot be integrable. Continue reading Counterexamples around Lebesgue’s Dominated Convergence Theorem

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 \[
f(t, \cdot) : & \Omega & \longrightarrow & \mathbb R \\
& \omega & \longmapsto & f(t,\omega) \end{array}
\] is integrable, one can define the function \[
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

A function continuous at all irrationals and discontinuous at all rationals

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:
\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

\(f\) 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

An unbounded positive continuous function with finite integral

Consider the piecewise linear function \(f\) defined on \([0,+\infty)\) taking following values for all \(n \in \mathbb{N^*}\):
0 & \mbox{if } x=0\\
0 & \mbox{if } x=n-\frac{1}{2n^3}\\
n & \mbox{if } x=n\\
0 & \mbox{if } x=n+\frac{1}{2n^3}\\

The graph of \(f\) can be visualized in the featured image of the post. Continue reading An unbounded positive continuous function with finite integral