# 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 $\begin{array}{l|rcl} f(t, \cdot) : & \Omega & \longrightarrow & \mathbb R \\ & \omega & \longmapsto & f(t,\omega) \end{array}$ is integrable, one can define the function $\begin{array}{l|rcl} F : & T & \longrightarrow & \mathbb R \\ & t & \longmapsto & \int_\Omega f(t,\omega) \ d\mu(\omega) \end{array}$

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 trigonometric series that is not a Fourier series (Lebesgue-integration)

We already provided here an example of a trigonometric series that is not the Fourier series of a Riemann-integrable function (namely the function $$\displaystyle x \mapsto \sum_{n=1}^\infty \frac{\sin nx}{\sqrt n}$$).

Applying an Abel-transformation (like mentioned in the link above), one can see that the function $f(x)=\sum_{n=2}^\infty \frac{\sin nx}{\ln n}$ is everywhere convergent. We now prove that $$f$$ cannot be the Fourier series of a Lebesgue-integrable function. The proof is based on the fact that for a $$2 \pi$$-periodic function $$g$$, Lebesgue-integrable on $$[0,2 \pi]$$, the sum $\sum_{n=1}^\infty \frac{c_n-c_{-n}}{n}$ is convergent where $$(c_n)_{n \in \mathbb Z}$$ are the complex Fourier coefficients of $$g$$: $c_n = \frac{1}{2 \pi} \int_0^{2 \pi} g(t)e^{-ikt} \ dt.$ As the series $$\displaystyle \sum_{n=2}^\infty \frac{1}{n \ln n}$$ is divergent, we will be able to conclude that the sequence defined by $\gamma_0=\gamma_1=\gamma_{-1} = 0, \, \gamma_n=- \gamma_{-n} = \frac{1}{\ln n} \ (n \ge 2)$ cannot be the Fourier coefficients of a Lebesgue-integrable function, hence that $$f$$ is not the Fourier series of any Lebesgue-integrable function. Continue reading A trigonometric series that is not a Fourier series (Lebesgue-integration)

# Intersection and sum of vector subspaces

Let’s consider a vector space $$E$$ over a field $$K$$. We’ll look at relations involving basic set operations and sum of subspaces. We denote by $$F, G$$ and $$H$$ subspaces of $$E$$.

### The relation $$(F \cap G) + H \subset (F+H) \cap (G + H)$$

This relation holds. The proof is quite simple. For any $$x \in (F \cap G) + H$$ there exists $$y \in F \cap G$$ and $$h \in H$$ such that $$x=y+h$$. As $$y \in F$$, $$x \in F+H$$ and by a similar argument $$x \in F+H$$. Therefore $$x \in (F+H) \cap (G + H)$$.

Is the inclusion $$(F \cap G) + H \subset (F+H) \cap (G + H)$$ always an equality? The answer is negative. Take for the space $$E$$ the real 3 dimensional space $$\mathbb R^3$$. And for the subspaces:

• $$H$$ the plane of equation $$z=0$$,
• $$F$$ the line of equations $$y = 0, \, x=z$$,
• $$G$$ the line of equations $$y = 0, \, x=-z$$,

# Playing with liminf and limsup

Let’s consider real sequences $$(a_n)_{n \in \mathbb N}$$ and $$(b_n)_{n \in \mathbb N}$$. We look at inequalities involving limit superior and limit inferior of those sequences. Following inequalities hold:
\begin{aligned} & \liminf a_n + \liminf b_n \le \liminf (a_n+b_n)\\ & \liminf (a_n+b_n) \le \liminf a_n + \limsup b_n\\ & \liminf a_n + \limsup b_n \le \limsup (a_n+b_n)\\ & \limsup (a_n+b_n) \le \limsup a_n + \limsup b_n \end{aligned} Let’s prove for example the first inequality, reminding first that $\liminf\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \left(\inf\limits_{m \ge n} a_m \right).$ For $$n \in \mathbb N$$, we have for all $$m \ge n$$ $\inf\limits_{k \ge n} a_k + \inf\limits_{k \ge n} b_k \le a_m + b_m$ hence $\inf\limits_{k \ge n} a_k + \inf\limits_{k \ge n} b_k \le \inf\limits_{k \ge n} \left(a_k+b_k \right)$ As the sequences $$(\inf\limits_{k \ge n} a_k)_{n \in \mathbb N}$$ and $$(\inf\limits_{k \ge n} b_k)_{n \in \mathbb N}$$ are non-increasing we get for all $$n \in \mathbb N$$, $\liminf a_n + \liminf b_n \le \inf\limits_{m \ge n} \left(a_m+b_m \right)$ which leads finally to the desired inequality $\liminf a_n + \liminf b_n \le \liminf (a_n+b_n).$ Continue reading Playing with liminf and limsup