Variations on Fatou’s Lemma – Part 1

November 13, 2014 Leave a comment

One of the most basic and powerful results of measure theory is Fatou’s Lemma. It provides information on the convergence of the integrals of functions which converge pointwise. The power of this lemma is the simplicity of its hypothesis. One of its interpretations is that the functional which to a positive function associates its integral is lower semicontinuous with respect to the pointwise convergence. Among other things, it allows us to prove the monotone convergence theorem, or Lebesgue’s dominated convergence theorem in just one line.

Fatou’s Lemma Suppose {(X,\mathcal{M},\mu)} is a measure space and {f_n : X \rightarrow [0,\infty]} is a sequence of integrable functions converging pointwise {\mu} almost everywhere to {f}. Then {f} is measurable and

\displaystyle \int_X f d\mu \leq \liminf_{n \rightarrow \infty} \int_X f_n d\mu.

Proof: One of the classical proofs of this result goes as follows. Using the definition

\displaystyle \int_X fd\mu = \sup\{ \int_X \varphi : \varphi \leq f,\ \varphi \text{ is a simple function} \}

we note that it is enough to prove that

\displaystyle \int_X \varphi d\mu \leq \liminf_{n \rightarrow \infty} \int_X f_n d\mu,

for every simple function {\varphi \leq f}.

Pick {\varphi\leq f} and {\varepsilon>0}. Suppose {\int_X \varphi d\mu<\infty}. Denote {E_n = \{x : (1-\varepsilon)\varphi \leq f_k(x),\ \text{ for every }k\geq n\}}. Note that {E_n \subset E_{n+1}} and {X = \bigcup_{n \geq 1} E_n}. If we denote {E} the support of {\varphi}, we may choose {n_0} such that for every {n \geq n_0} we have {\mu(E\setminus E_n)<\varepsilon}. Thus

\displaystyle \int_X f_k d\mu_k \geq \int_{E_k} f_k d\mu \geq (1-\varepsilon)\int_{E_k}\varphi d \mu \geq (1-\varepsilon) \int_X \varphi d\mu - (1-\varepsilon) \int_{E\setminus E_k} \varphi d\mu.

Since {\varphi} is bounded above by a constant {M<\infty} and {\mu(E\setminus E_k)<\varepsilon}, we have

\displaystyle \int_X f_k d\mu \geq (1-\varepsilon) \int_E \varphi d\mu -(1-\varepsilon)\varepsilon M.

Taking {k \rightarrow \infty} we have

\displaystyle \liminf_{k \rightarrow \infty} \int_X f_k d\mu \geq (1-\varepsilon) \int_E \varphi d\mu -(1-\varepsilon)\varepsilon M.

Take {\varepsilon \rightarrow 0} and then supremum over {\varphi\leq f} to finish the proof. \hfill {\square}

Monotone convergence Suppose that {f_k} is a sequence of function defined on {X} such that

\displaystyle 0 \leq f_k(x) \leq f_{k+1}(x),

for every {k \geq 1} and {x \in X}. Denote {f(x) = \lim_{k \rightarrow \infty} f_k(x)}. Then {f} is measurable and

\displaystyle \int_X fd\mu = \lim_{n \rightarrow \infty} \int_X f_n d\mu.

Proof: The {\limsup} comes from the monotonicity, while the {\liminf} is just Fatou’s Lemma.

Dominated convergence Suppose {f_k} is a sequence of measurable functions defined on {X} which converges almost everywhere to {f}. Moreover, suppose there exists an integrable function {g} such that {|f_n|\leq g} for every {n}. Then

\displaystyle \int_X fd\mu = \lim_{n \rightarrow \infty} \int_X f_n d\mu.

Proof: Apply Fatou’s lemma to {g_n = 2g-|f_n-f|\geq 0}.

It is straightforward to see that the positivity condition can be replaced with a domination from below. If {f_n \geq -g} and {g} is integrable, then apply Fatou’s lemma to {f_n+g} to deduce that the conclusion still holds for {(f_n)}. One may ask what happens with the {\limsup} part. If { f_n \leq g} with {g} integrable, then the reverse Fatou Lemma works. To see this, apply Fatou’s lemma to {g-f_n \geq 0}.

Of course, every one of the few hypotheses made for the Fatou’s lemma are essential. If we allow negative functions, then {f_n = -n \chi_{[0,1/n]}} is a counterexample. {f_n} converges to zero almost everywhere, while the integrals of {f_n} are equal to {-1}. The inequality can be strict. Consider the similar functions {f_n = n \chi_{[0,1/n]}}.

Seeing that the hypotheses needed for Fatou lemma to hold are simple enough, we may try and see which generalizations of Fatou lemma hold? One interesting direction is to consider not only variable functions, but also variable measures. This will be treated in a future post.

Problems of the Miklos Schweitzer Competition 2014

November 10, 2014 Leave a comment

Problem 1. Let {n} be a positive integer. Let {\mathcal{F}} be a familiy of sets that contains more than half of all subsets of an {n}-element set {X}. Prove that from {\mathcal{F}} we can select {\lceil \log_2 n\rceil+1 } sets that form a separating family of {X}, i.e., for any two distinct elements of {X} there is a selected set containing exactly one of the two elements.

Problem 2. let {k \geq 1} and let {I_1,...,I_k} be non-degenerate subintervals of the interval {[0,1]}. Prove that

\displaystyle \sum \frac{1}{|I_i \cup I_j|} \geq k^2,

where the summation is over all pairs of indices {(i,j)} such that {I_i} and {I_j} are not disjoint.

Problem 3. We have {4n+5} points in the plane, no three of them collinear. The points are colored with two colors. Prove that from the points we can form {n} empty triangles (they have no colored points in their interiors) with pairwise disjoint interiors, such that all points occuring as vertices of the {n} triangles have the same color.

Problem 4. For a positive integer {n}, let {f(n)} be the number of sequences {a_1,...,a_k} of positive integers such that {a_i \geq 2} and {a_1...a_k = n} for {k \geq 1}. We make the convention {f(1)=1}. Let {\alpha} be the unique real number greater than {1} such that {\sum_{n=1}^\infty n^{-\alpha}=2}. Prove that

  • (i) { \sum_{ k = 1}^n f(k)= O(n^\alpha)}.
  • (ii) There exists no number {\beta<\alpha} such that {f(n)=O(n^\beta)}.

Problem 5. Let {\alpha} be a non-real algebraic integer of degree two, and let {P} be the set of irreducible elements of the ring {\Bbb{Z}[\alpha]}. Prove that

\displaystyle \sum_{ p \in P} \frac{1}{|p|^2} = \infty.

Problem 6. Let {\rho : G \rightarrow GL(V)} be a representation of a finite {p}-group {G} over a field of characteristic {p}. Prove that if the restriction of the linear map {\sum_{ g \in G} \rho(g)} to a finite dimensional subspace {W} of {V} is injective, then the subspace spanned by the subspaces {\rho(g)W} ({g \in G}) is the direct sum of these subspaces.

Problem 7. Lef {f: \Bbb{R} \rightarrow \Bbb{R}} be a continuous function and let {g: \Bbb{R} \rightarrow \Bbb{R}} be arbitrary. Suppose that the Minkowski sum of the graph of {f} and the graph of {g} (i.e. the set {\{(x+y,f(x)+g(y) : x,y \in \Bbb{R}\}} has Lebesgue measure zero. Does it follow then that the function {f} is of the form {f(x)=ax+b}, with suitable constants {a,b \in \Bbb{R}}?

Problem 8. Let {n \geq 1} be a fixed integer. Calculate the distance

\displaystyle \inf_{p,f} \max_{0 \leq x \leq 1} |f(x)-p(x)|,

where {p} runs over polynomials of degree less than {n} with real coefficients and {f} runs over functions of the form

\displaystyle f(x) = \sum_{ k = n}^\infty c_kx^k

defined on the closed interval {[0,1]}, where {c_k\geq 0} and {\sum_{k=n}^\infty c_k =1}.

Problem 9. Let {\rho : \Bbb{R}^n \rightarrow \Bbb{R},\ \rho(x)=e^{-\|x\|^2}}, and let {K \subset \Bbb{R}^n} be a convex body, i.e. a compact convex set with nonempty interior. Define the barycenter {s_K} of the body {K} with respect to the weight function {\rho} by the usual formula

\displaystyle s_K = \frac{\int_K \rho(x) x dx}{\int_K \rho(x)dx}.

Prove that the translates of the body {K} have pairwise distinct barycenters with respect to {\rho}.

Problem 10. To each vertex of a given triangulation of the two dimensional sphere, we assign a convex subset of the plane. Assume that the three convex sets corresponding to the three vertices of any two dimensional face of the triangulation have at least one point in common. Show that there exist four vertices such that the corresponding convex sets have at least one point in common.

Problem 11. Let {U} be a random variable that is uniformly distributed on the interval {[0,1]}, and let

\displaystyle S_n = 2\sum_{k=1}^n \sin(2kU\pi).

Show that, as {n \rightarrow \infty}, the limit distribution of {S_n} is the Cauchy distribution with density function {f(x) =\frac{1}{\pi(1+x^2)}}.

Source: http://www.bolyai.hu/SCH_angol_2014.pdf

Monotonic bijection from naturals to pairs of natural numbers

October 23, 2014 Leave a comment

This is a cute problem I found this evening.

Suppose {\phi : \Bbb{N}^* \rightarrow \Bbb{N}^*\times \Bbb{N}^*} is a bijection such that if {\phi(k) = (i,j),\ \phi(k')=(i',j')} and {k \leq k'}, then {ij \leq i'j'}.

Prove that if {k = \phi(i,j)} then {k \geq ij}.

Proof: The trick is to divide the pairs of positive integers into families with the same product.

\displaystyle \begin{matrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) & \cdots \\ & (2,1) & (3,1) & (2,2) & (5,1) & (2,3) & \cdots \\ & & &( 4,1) & & (3,2) & \cdots \\ & & & & & (6,1) & \cdots \end{matrix}

Note that the {M}-th column contains as many elements as the number of divisors of {M}. Now we just just use a simple observation. Let {\phi(k)=(i,j)} be on the {M}-th column (i.e. {ij = M}). If {n \geq 1} then {\phi(k+n)=(i',j')} cannot be on one of the first {M-1} columns. Indeed, the monotonicity property implies {M = ij \leq i'j'}. The fact that {\phi} is a bijection assures us that {\phi(1),...,\phi(k)} cover the first {M-1} columns. Moreover, one element from the {M}-th column is surely covered, namely {(i,j) = \phi(k)}. This means that

\displaystyle k \geq d(1)+...+d(M-1)+1 \geq M = ij,

where we have denoted by {d(n)} the number of positive divisors of {n}.

Solving Poisson’s equation on a general shape using finite differences

October 14, 2014 3 comments

One of the questions I received in the comments to my old post on solving Laplace equation (in fact this is Poisson’s equation) using finite differences was how to apply this procedure on arbitrary domains. It is not possible to use this method directly on general domains. The main problem is the fact that, unlike in the case of a square of rectangular domain, when we have a general shape, the boudary can have any orientation, not only the orientation of the coordinate axes. One way to avoid approach this problem would be using the Finite Element Method. Briefly, you discretize the boundary, you consider a triangulation of the domain with respect to this discretization, then you consider functions which are polynomial and have support in a few number of triangles. Thus the problem is reduced to a finite dimensional one, which can be written as a matrix problem. The implementation is not straightforward, since you need to conceive algorithms for doing the discretization and triangulation of your domain.

One other approach is to consider a rectangle {D} which contains the shape {\Omega} and add a penalization on the exterior of your domain {\Omega}. The problem to solve becomes something like:

\displaystyle (-\Delta +\mu I) u = 1 \text{ on }D

where {\mu} is defined by

\displaystyle \mu(x) = \begin{cases} 0 & x \in \Omega \\ + \infty & x \notin \Omega\end{cases}. \ \ \ \ \ (1)

Note that doing this we do not need to impose the boundary condition {u=0} on {\partial \Omega}. This is already imposed by {\mu}, and the fact that {u} is forced to be zero outside {\Omega}.

Read more…

Generalized version of Lebesgue dominated convergence theorem

October 10, 2014 Leave a comment

The following variant of the Lebesgue dominated convergence theorem may be useful in the case we can not dominate a sequence of functions by only one integrable function, but by a convergent sequence of integrable functions.

Let {(f_n)} be a sequence of measurable functions defined on a measurable set {\Omega} with real values, which converges pointwise almost everywhere to {f}. Suppose that there exists a sequence of positive, integrable functions {g_n} such that the two following conditions hold:

  • {|f_n(x)|\leq g_n(x)} for almost every {x \in \Omega};
  • There exists an integrable function {g} such that {(g_n) \rightarrow g} in {L^1(\Omega)}.

Then

\displaystyle \lim_{n \rightarrow \infty} \int_\Omega f_n = \int_\Omega f

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Necessary condition for the uniform convergence of a certain series

October 8, 2014 Leave a comment

Let {a_n} be a decreasing sequence of positive numbers such that the series

\displaystyle \sum_{n = 1}^\infty a_n \sin(nx)

is uniformly convergent. Then {(a_n)} must satisfy {(na_n) \rightarrow 0}.

Note that this result implies that the series {\sum_{n=1}^\infty a_n \sin(nx)} is not uniformly convergent on {\Bbb{R}}. It is surprisngly similar to the following result:

Suppose that {(a_n)} is a decreasing sequence of positive real numbers such that the series {\sum_{n=1}^\infty a_n} is convergence. Then {(na_n) \rightarrow 0}. It is no surprise that the proofs of these two results are similar.

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Shrinkable polygons

October 7, 2014 Leave a comment

Here’s a nice problem inspired from a post on MathOverflow: link

We call a polygon shrinkable if any scaling of itself with a factor {0<\lambda<1} can be translated into itself. Characterize all shrinkable polygons.

It is easy to see that any star-convex polygon is shrinkable. Pick the point {x_0} in the definition of star-convex, and any contraction of the polygon by a homothety of center {x_0} lies inside the polygon.

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