Posts Tagged ‘integral’

IMC 2016 – Day 2 – Problem 7

July 28, 2016 Leave a comment

Problem 7. Today, Ivan the Confessor prefers continuous functions {f:[0,1]\rightarrow \Bbb{R}} satisfying {f(x)+f(y) \geq |x-y|} for all {x,y \in [0,1]}. Fin the minimum of {\int_0^1 f} over all preferred functions.

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SEEMOUS 2016 Problem 4 – Solution

March 6, 2016 3 comments

Problem 4. Let {n \geq 1} be an integer and set

\displaystyle I_n = \int_0^\infty \frac{\arctan x}{(1+x^2)^n}dx.

Prove that

a) {\displaystyle \sum_{i=1}^\infty \frac{I_n}{n} =\frac{\pi^2}{6}.}

b) {\displaystyle \int_0^\infty \arctan x \cdot \ln \left( 1+\frac{1}{x^2}\right) dx = \frac{\pi^2}{6}}.

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Variations on Fatou’s Lemma – Part 2

November 29, 2014 Leave a comment

As we have seen in a previous post, Fatou’s lemma is a result of measure theory, which is strong for the simplicity of its hypotheses. There are cases in which we would like to study the convergence or lower semicontinuity for integrals of the type {\displaystyle \int_\Omega f_ng_n} where {f_n} converges pointwise to {f} and {g_n} converges to {g} in some fashion, but not pointwise. For example, we could have that {g_n} converges to {g} in {L^1}. In this case we could write the integral {\displaystyle \int_\Omega f_ng_n} as {\displaystyle \int_\Omega f d\mu_n} where {\mu_n} is the measure defined by {\mu_n(A) = \displaystyle\int_A g_n}. All measures considered in this post will be positive measures.

Certain hypotheses on the measures {\mu_n,\mu} allow us to find a result similar to Fatou’s Lemma for varying measures. In the following, we define a type of convergence for the measures {\mu_n,\mu}, named setwise convergence, which will allow us to prove the lower semicontinuity result. We say that {\mu_n} converges setwise to {\mu} if {\mu_n(A) \rightarrow \mu(A)} for every measurable set {A}. The following proof is taken from Royden, H.L., Real Analysis, Chapter 11, Section 4. It is very similar to the proof of Fatou’s lemma given here.

Theorem A. Let {\mu_n} be a sequence of measures defined on {\Omega} which converges setwise to a measure {\mu} and {(f_n)} a sequence of nonnegative measurable functions which converge pointwise (or almost everywhere in {\Omega}) to the function {f}. Then

\displaystyle \int_\Omega f d\mu \leq \liminf_{n \rightarrow \infty} \int_\Omega f_n d\mu_n.

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Variations on Fatou’s Lemma – Part 1

November 13, 2014 1 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.

Generalized version of Lebesgue dominated convergence theorem

October 10, 2014 1 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)}.


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

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March 18, 2014 Leave a comment

Problem 1. Let {n} be a nonzero natural number and {f:\Bbb{R} \rightarrow \Bbb{R}\setminus \{0\}} be a function such that {f(2014)=1-f(2013)}. Let {x_1,..,x_n} be distinct real numbers. If

\displaystyle \left| \begin{matrix} 1+f(x_1)& f(x_2)&f(x_3) & \cdots & f(x_n) \\ f(x_1) & 1+f(x_2) & f(x_3) & \cdots & f(x_n)\\ f(x_1) & f(x_2) &1+f(x_3) & \cdots & f(x_n) \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ f(x_1)& f(x_2) & f(x_3) & \cdots & 1+f(x_n) \end{matrix} \right|=0

prove that {f} is not continuous.

Problem 2. Consider the sequence {(x_n)} given by

\displaystyle x_1=2,\ \ x_{n+1}= \frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2},\ n \geq 2.

Prove that the sequence {y_n = \displaystyle \sum_{k=1}^n \frac{1}{x_k^2-1} ,\ n \geq 1} is convergent and find its limit.

Problem 3. Let {A \in \mathcal{M}_n (\Bbb{C})} and {a \in \Bbb{C},\ a \neq 0} such that {A-A^* =2aI_n}, where {A^* = (\overline A)^t} and {\overline A} is the conjugate matrix of {A}.

(a) Show that {|\det(A)| \geq |a|^n}.

(b) Show that if {|\det(A)|=|a|^n} then {A=aI_n}.

Problem 4. a) Prove that {\displaystyle \lim_{n \rightarrow \infty} n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx=\frac{\pi}{2}}.

b) Find the limit {\displaystyle \lim_{n \rightarrow \infty} n\left(n \int_0^n \frac{\arctan(x/n)}{x(x^2+1)}dx-\frac{\pi}{2} \right)}

SEEMOUS 2012 Problem 4

March 8, 2012 Leave a comment

a) Compute \displaystyle \lim_{n \to \infty} n \int_0^1 \left(\frac{1-x}{1+x} \right)^n dx.

b) Let k \geq 1 be an integer. Compute

\displaystyle\lim_{n \to \infty}n^{k+1}\int_0^1 \left( \frac{1-x}{1+x}\right)^n x^k dx.

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