What is a Semigroup and How Does it Relate to Immeasurable Sets?

[simpleton1]

Hi,

I'm taking a course in functional analysis and having some trouble with the following questions :

1. L1(R) is the space of absolutely integrable functions on R with the norm integrate(abs(f(x)) over -inf to +inf.
Define a linear operator from L1(R) to L1(R) as A(f)(x)=integrate (f(x-y)/(1+y^2))dy over -inf to +inf.
prove the operator is bound and find its norm.
I've integrated norm(A(f)(x)) to a double integral pi*integrate(integrate(abs(f(x-y))dydx over inf to +inf for both
integrals. I don't know how to proceed to show that it's a product of a constant and norm(f(x)).

2. Find the inverse operator for A(f)(x) = f(x)-0.5integrate(f(t)dt) from 0 to x. The operator is defined from C[0,1] to
C[0,1] where C[0,1] are the continuos functions in [0,1] with the supremom norm.
They say the inverse operator is of the form A^-1=f(x)+integrate(f(t)K(x-t)dt) from 0 to x.
previous sub questions which are to be used for solving this say that if A is a bound operator whose norm is
smaller than 1 then if B=I-A it is inversible and its inverse is B^-1=I+A+A^2...
another sub questions says the solution for differential equation g[n](x)=f(x),g[n-1](0)=...g'(0)=0
is 1/(n-1)!*integrate(f(t)(x-t)^n-1) from 0 to x.

3. Prove there one and only continuous function that solves the equation :
f(x) = sinx + integrate(f(y)exp(-x-y-1)dy over 0 to 1, x belongs to [0,1].
I've solved the equation and found the function by guessing a solution but don't know how to prove it's the only
solution.

Any help will be much appreciated.

 

[Euge]

Hello simpleton,

Welcome! (Wave) Since this is your first post, let me explain how you can improve your post with proper typesetting. Review the http://mathhelpboards.com/latex-tips-tutorials-56/mhb-latex-guide-pdf-1142.html to learn how to format equations and (particularly in your case) typeset integrals. Instead of writing

integrate(f(x-y)/(1 + y^2) dy) from -inf to inf

you can enter \$\$\int_{-\infty}^\infty \frac{f(x-y)}{1+y^2}\, dy\$\$, which produces

$$\int_{-\infty}^\infty \frac{f(x-y)}{1+y^2}\, dy.$$

Notice the dollar signs that wrap the integral expression. If you use single dollar signs, such as \$f(x)\$, you'll produce a math display $f(x)$ within the text, as you can see. But if you use two dollar signs, such as \$\$f(x)\$\$, then it'll result in a display line separate from text, like this:

$$f(x)$$

I hope these tips are clear to you. Now back to your questions.

1. Let $f\in \mathcal{L}^1(\Bbb R)$. By Minkowski's inequality,

$$\|A(f)\|_{\mathcal{L}^1(\Bbb R)} \le \int_{-\infty}^\infty \left\|\frac{f(x - y)}{1+y^2}\right\|_{\mathcal{L}_x^1(\Bbb R)}\, dy = \int_{-\infty}^\infty \frac{\|f(x-y)\|_{\mathcal{L}^1_x(\Bbb R)}}{1+y^2}\, dy \overset{(*)}{=} \int_{-\infty}^\infty \frac{\|f\|_{\mathcal{L}^1(\Bbb R)}}{1 + y^2}\, dy = \pi \|f\|_{\mathcal{L}^1(\Bbb R)}.$$

The step $(*)$ is justified by the translation invariance of the Lebesgue measure on $\Bbb R$. Thus $\|A\| \le \pi$. To see that $\|A\| = \pi$, consider the function $f(x) = \frac{1}{\sqrt{\pi}}e^{-x^2}$, $-\infty < x < \infty$. Show that for this choice of $f$, $\|f\|_{\mathcal{L}^1(\Bbb R)} = 1$ and $\|A(f)\|_{\mathcal{L}^1(\Bbb R)} = \pi$. For this proves $\|A\| \ge \pi$, and hence $\|A\| = \pi$.

Remark. In case you're not aware of Minkowski's integral inequality, you can write this alternative: since for each $x\in \Bbb R$, $\lvert A(f)(x)\rvert \le \int_{-\infty}^\infty \frac{|f(x-y)|}{1+y^2}\, dy$, then

$$\|A(f)\|_{\mathcal{L}^1(\Bbb R)} \le \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\lvert f(x-y)\rvert}{1 + y^2}\, dy\, dx = \int_{-\infty}^\infty \int_{-\infty}^\infty \lvert f(x-y)\rvert\, dx\, \frac{dy}{1+y^2},$$

using the Fubini-Tonelli theorem in the last step; by translation invariance

$$\int_{-\infty}^\infty \lvert f(x-y)\rvert\, dx = \int_{-\infty}^\infty \lvert f(x)\rvert\, dx = \|f\|_{\mathcal{L}^1(\Bbb R)},$$

so indeed $\|A(f)\|_{\mathcal{L}^1(\Bbb R)} \le \pi \|f\|_{\mathcal{L}^1(\Bbb R)}$.
____________________________________________________________________________________________________2. Let $f\in C[0,1]$. We want to solve the equation $A(g) = f$ for $g$. To do so, let

$$G(x) := \int_0^x g(t)\, dt \qquad (x\in [0,1]).$$

since $g\in C[0,1]$, $G$ is differentiable by the FTC (I'm using an abbreviation for the fundamental theorem of calculus) with

$$G'(x) = g(x) = A(g)(x) + 0.5\int_0^x g(t)\, dt = f(x) + 0.5G(x).$$

Since $G(0) = 0$, we have an IVP

$$G' - 0.5G = f,\, G(0) = 0.$$

An integrating factor for this ODE is $e^{-0.5x}$, so

$$(e^{-0.5x}G)' = f(x)e^{-0.5x}.$$

Therefore, since $G(0) = 0$,

$$e^{-0.5x}G(x) = \int_0^x f(t)e^{-0.5t}\, dt,$$

or

$$G(x) = \int_0^x f(t) e^{0.5(x-t)}\, dt.$$

Differentiating with respect to $x$, using FTC and the Leibniz rule we obtain

$$g(x) = f(x)e^{0.5(x-x)} + \int_0^x \frac{\partial}{\partial x}[f(t)e^{0.5(x-t)}]\, dt = f(x) + 0.5\int_0^x f(t)e^{0.5(x-t)}\, dt.$$

Hence, the inverse of $A$ is given by

$$A^{-1}(f)(x) = f(x) + 0.5\int_0^x f(t)e^{0.5(x-t)}\, dt \qquad (f\in C[0,1], x\in [0,1]).$$
____________________________________________________________________________________________________

3. It's great that you have found an explicit solution to the integral equation, but this question is a qualitative question regarding existence and uniqueness -- the key here is to apply Banach's contraction principle (or, as some call it, Banach's fixed point theorem).

Let's recall Banach's contraction principle. Suppose $(M,d)$ is a metric space. A contraction mapping on $M$ is a mapping $f : M \to M$ such that there is some $c\in (0,1)$ such that for all $x,y\in M$,

$$d(f(x),f(y)) \le c\, d(x,y).$$

The number $c$ is known as the contraction constant. Banach's contraction principle states that if $(M,d)$ is a complete metric space and $f$ is a contraction mapping on $M$, then $f$ has a unique fixed point.

To set up the Banach contraction scheme, define an operator $A : C[0,1] \to C[0,1]$ by the equation

$$A(f)(x) = \sin x + \int_0^1 f(y)e^{-x-y-1}\, dy.$$

The metric $d$ we'll use here is the one induced from the max norm, i.e.,

$$d(f,g) = \max\{\lvert f(x) - g(x)\rvert :x\in [0,1]\}.$$

For all $f, g\in C[0,1]$,

$$d(A(f),A(g)) \le \int_0^1 \lvert f(y) - g(y)\rvert e^{-y-1}\, dy \le e^{-1}d(f,g),$$

which means that $A$ is a contraction mapping on $C[0,1]$ with contraction contraction constant $e^{-1}$. As $(C[0,1],d)$ is complete, by Banach's contraction principle, there is a unique $f\in C[0,1]$ such that $A(f) = f$. Hence, your integral equation has a unique solution in $C[0,1]$.

 

[simpleton1]

Thank you very much!

Since you've been so helpful I want to ask two additional, simpler questions (hope I got the notation right) :

1. $$\left\lVert{P}\right\rVert = (\sum_{n=0}^{\infty} {P}^{2} (\left(\frac{1}{2}\right)^{\!{n}}))^\frac{1}{2}$$
where P is a polynomial defined over [0,1] and P(0)=0.
The question was to prove that P is finite in [0,1] and that it is a norm. I've done everything except prove
the traingle inequality - it seems to not hold because
$$\left\lVert{P+Q}\right\rVert = (\sum_{n=0}^{\infty} {(P+Q)}^{2} (\left(\frac{1}{2}\right)^{\!{n}}))^\frac{1}{2} $$

2. E is an immeasurable group in (0,1) . Define function f = xI_E + x^3I_Ec
where I_E is the indicating function for E and I_Ec is the indicating function for E's complement.
a. Show that f's inverse is measurable over R (I've done it but not sure that correctly).
b. Is f measurable?

 

[Euge]

1. So you've defined, for all polynomials $P$ over $[0,1]$ with $P(0) = 0$,

$$\|P\| = \left[\sum_{n = 0}^\infty P^2\left(\frac{1}{2^n}\right)\right]^{1/2}.$$

Before discussing the triangle inequality, it's important to note why this norm makes sense, i.e., why the series above is convergent.

Suppose $P$ has degree $m > 1$, and let $c_1,\ldots, c_m$ be scalars such that for all $x\in [0,1]$,

$$P(x) = c_1x + c_2x^2 + \cdots + c_m x^m$$

(This can be done since $P(0) = 0$). Then for all $x\in [0,1]$,

$$|P(x)| = \lvert c_1 + c_2 x + \cdots + c_m x^{m-1}\rvert\, \lvert x \rvert \le C\lvert x\rvert,$$

where $C = |c_1| + \cdots + |c_m|$. Therefore

$$ P^2\left(\frac{1}{2^n}\right) \le \frac{C^2}{2^{2n}} \quad (n = 0, 1, 2, \ldots).$$

Since the geometric series $\sum_{n = 0}^\infty \frac{C^2}{2^{2n}}$ converges, then by direct comparison, the series $\sum_{n = 0}^\infty P^2(1/2^n)$ converges. Thus, $\|P\|$ is finite (in fact, bounded above by $2C/\sqrt{3}$).

Let $\Bbb N_0$ denote the set of whole numbers. Since

$$\|P\| = \left\|P\left(\frac{1}{2^n}\right)\right\|_{\ell^2(\Bbb N_0)}$$

and the triangle inequality holds for the $\ell^2$-norm on $\Bbb N_0$, then the triangle inequality holds for $\|\cdot \|$.Show me what you've done so far in part 2. so I can make comments.

 

[simpleton1]

Thanks again for your help.

About 1 - I'm still not sure how it can be that the triangle inequality holds since
$$(\sum_{n=0}^{\infty}{(P+Q)}^{2}\left(\frac{1}{2}\right)^{\!{n}})^{1/2}>\left\lVert{P+Q}\right\rVert$$
for polynomes.

About 2 - I didn't write the question in part a correctly. It was to prove that for every y$\in$R
${f}^{-1}(y)$ is measureable. I used the fact that for every y ${f}^{-1}(y)$ can receive between
0 and 2 values and since this is a finite group for every y it is also measureable.
for part b - I used the fact the ${I}_{{E}^{c}}= 1 - {I}_{E}$
so that $f(x) = x({I}_{E}(x)(1-{x}^{2})+1)$.
In order for f to be measureable ${f}^{-1}$ needs to transfer every Borel set to a Borel set.
That's where I got stuck since I couldn't prove that it does or doesn't do it.

 

[Euge]

The triangle inequality for $\mathcal{L}^p$-norms is known as Minkowski's inequality. In case you're not familiar with it, I'll prove from scratch that $\|\cdot \|$ satisfies the triangle inequality. To do so, I'll use the inequality

$$(x + y)^2 \le \lvert x\rvert \, \lvert x + y\rvert + \lvert y \rvert \, \lvert x + y\rvert$$

over the real numbers. It follows from the fact that $(x + y)^2 = |x + y|^2 = \lvert x + y\rvert \, \lvert x + y\rvert$ and $\lvert x + y\rvert \le \lvert x\rvert + \lvert y\rvert$.

Given polynomials $P, Q$ over $[0,1]$ with $P(0) = 0 = Q(0)$,

$$\|P + Q\|^2 = \sum_{n = 0}^\infty \left[P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right]^2 \le \sum_{n = 0}^\infty \left\lvert P\left(\frac{1}{2^n}\right)\right\rvert\, \left\lvert P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right\rvert + \sum_{n = 0}^\infty \left\lvert Q\left(\frac{1}{2^n}\right)\right\rvert\, \left\lvert P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right\rvert.\tag{1}$$

By the Cauchy-Schwarz inequality,

$$\sum_{n = 0}^\infty \left\lvert P\left(\frac{1}{2^n}\right)\right\rvert\, \left\lvert P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right\rvert \le \left[\sum_{n = 0}^\infty P^2\left(\frac{1}{2^n}\right)\right]^{1/2}\left\{\sum_{n = 0}^\infty \left[P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right]^2\right\}^{1/2} = \|P\|\, \|P + Q\|\tag{2}$$

and similarly

$$\sum_{n = 0}^\infty \left\lvert Q\left(\frac{1}{2^n}\right)\right\rvert \, \left\lvert P\left(\frac{1}{2^n}\right) + Q\left(\frac{1}{2^n}\right)\right\rvert \le \|Q\|\, \|P + Q\|.\tag{3}$$

Combining $(1), (2)$, and $(3)$,

$$\|P + Q\|^2 \le \|P\|\, \|P + Q\| + \|Q\|\, \|P + Q\| = (\|P\| + \|Q\|) \|P + Q\|.\tag{4}$$

If $\|P + Q\| = 0$, then the inequality $\|P + Q\| \le \|P\| + \|Q\|$ holds trivially, so suppose $\|P + Q\| > 0$. Dividing $(4)$ by $\|P + Q\|$, the triangle inequality is obtained.

 

[Euge]

With Question #2, is $E$ a group under multiplication or addition?

 

[simpleton1]

All it says is that E is an immeasureable group contained in (0,1).

 

[Euge]

Check in your text or with your professor for clarification of an immeasurable group. Just keep in mind, $E$ cannot be a group under usual multiplication or addition in $\Bbb R$, because closure under inverses would not be satisfied. The problem would make more sense if $E$ is an immeasurable semigroup in $(0,1)$.

 

[simpleton1]

Perhaps what is missing is that in this course when they say immeasureable it means
immeasureable by the lebesgue measure.
I would copy the question text directly but it's not in english so I just translated it.

 

[Euge]

That part is fine, but what about the group part? What is the operation on $E$?

 

[simpleton1]

I tried to ask the professor but he also didn't understand the question.
We didn't learn about semi-groups in this course.
Could you please try and solve under reasonable assumptions?

 

[simpleton1]

Also might help me solve this is if the compliment of an immeasureable group is measureable.
Or if the combination of a measureable or immeasureable function is immeasureable.
Can any of these be proven?

 

[Euge]

Hi simpleton,

I think that if the professor does not understand the question, then the question should not be assigned. However, I will say this: A semigroup is a set with an associative binary operation. Since $E$ is immeasurable, $E$ is nonempty. Assuming the binary operation on $E$ is multiplication of real numbers, then for all $y\in \Bbb R$,

$$f^{-1}\{y\} = \begin{cases}\{y\},& y\in E\\ \{\sqrt[3]{y}\},&y\notin E\end{cases}$$

Since one-point sets in $\Bbb R$ are measurable (since they are closed), then for all $y\in \Bbb R$, $f^{-1}\{y\}$ is measurable.