Help for test - functional analysis

[simpleton1]

Hi - my professor in functional analysis posted 4 prior years tests just 4 days before the test without solutions.
I'd appreciate if anyone can help send solutions for the following with the following questions :

1. $\mu$ is a sigma additive measure over sigma algebra $\Sigma$.
A $\in \varSigma$ is some finite set and we'll define the function m:$\Sigma$ to R
such that $m(B) = \mu (A \cup B)$.
Show that m is a sigma additive measure iff $\mu (A) = 0$.

2. Let f:[0,$\infty$) to [0,$\infty$) transformation given by f(x)=x+exp(-x)
Prove that $\left| f(x)-f(y) \right| < \left| x-y \right|$ for every $x\ne y$.
Is there a constant a between 0 and 1 such that $\left| f(x)-f(y) \right| < a\left| x-y \right|$
for every $x\ne y$ at [0,$\infty$)

3. X is the sapce of all infinite series of real numbers with a finite number of
elements which are different than 0 (mark elements by (a1,a2,...)).
Define two norms over X :
$\left\lVert{a}\right\rVert{}_1{}$ = $\sum_{n}^{\infty} \left| a{}_n{} \right|$
and
$\left\lVert{a}\right\rVert{}_\infty{}$ = $max( \left| a{}_n{} \right| )$

Define operator L so that L (from X to X) shifts elements left and divides by their location :
(a1,a2,...) change to (a2/1,a2/3,a3/4,...).
Is L an obstructed linear transformation from X1 to X1? If yes - what is it's Norm?
Is L and obstrcuted linear transofrmation from $\left\lVert{X}\right\rVert{}_\infty{}$ to $\left\lVert{X}\right\rVert{}_1$

 

[Euge]

Re: help for test - functional analysis\mu

Hello again, simpleton,

Please edit your post, as it contains many errors.

1. The term "group" should be replaced with "measurable set".

2. The conclusion does not make sense. Did you mean "There exists a constant $\alpha$ such that for all $x,y\in [0,\infty)$, $x\neq y$ implies $|f(x) - f(y)| < \alpha|x - y|$?"

3. You've defined $\|a\|_1$ and $\|a\|_\infty$ exactly the same. It should be $\|a\|_\infty = \sup\{|a_n| :n\in \Bbb N\}$. The definition of $L$ seems off -- shouldn't it be $L(a) = \left(\frac{a_2}{1}, \frac{a_3}{2}, \frac{a_4}{3},\ldots\right)$?

Also, please put some space between problems so that it's easier to read.

 

[simpleton1]

Re: help for test - functional analysis\mu

Hi Euge - thanks for the commentation. I did post the questions under the influence of tiredness. :D

I've fixed most things I think. You are right about the definition of L in question 3. I just got too lazy to
Latex it.

 

[Euge]

1. It follows from the fact that $m(\emptyset) = 0$ if and only if $\mu(A) = 0$.

2. Note that $f'(x) = 1 - \exp(-x) < 1$ for all $x > 0$. Now use the mean value theorem to establish the inequality $\lvert f(x) - f(y)\rvert < \lvert x - y\rvert$ for all $x,y\in [0,\infty)$ with $x\neq y$. For the second statement, the answer is no. For otherwise $f'$ is bounded by $\alpha$ on $[0,\infty)$. Then $\sup\{f'(x) : x\in [0,\infty)\} = 1 > \alpha$.

3. If my definition of $L$ is correct (which still differs from yours), then $L$ is a linear operator on $X$. For all $a\in X_1$,

$$\|L(a)\|_1 = \sum_{n = 2}^\infty \frac{\lvert a_n\rvert}{n-1} \le \sum_{n = 2}^\infty \lvert a_n\rvert \le \|a\|_1,$$

thus $\|L\|_{X_1\to X_1} \le 1$. On the other hand, $\|(0,1,0,0,\ldots)\|_1 = 1$ and $\|L(0,1,0,0,\ldots)\|_1 = \|(1,0,0,0,\ldots)\|_1 = 1$. Therefore, $\|L\|_{X_1\to X_1} \ge 1$, proving that $\|L\|_{X_1\to X_1} = 1$.

However, $L$ is not a transformation from $X_\infty$ to $X_1$, since $a = (1,1,1,1,\ldots) \in X_\infty$ (with $\|a\|_\infty = 1$), but $\|L(a)\|_1 = \sum\limits_{n = 1}^\infty \frac{1}{n} = +\infty$.

 

[simpleton1]

Ok, Thanks.
I got the first two questions and the first part of question 3.
But the second part of 3 has still got me confused - since the number of elements in a different
than 0 are finite , the sum of $\left\lVert{a}\right\rVert\infty$ over a=(1,1,1,1,...0,0,0) will not
be $\infty$ but a finite number.

 

[Euge]

I was going to correct the error but you beat me to it! Actually, I defined $a$ to be the sequence of $1$'s, no zeros are in the sequence. That is a problem since then $a$ does not belong to $X$.

Let's try this again...

$L$ is not a transformation from $X_\infty$ to $X_1$. For given $m\in \Bbb N$, let $a^m = (1,1,\ldots,\underbrace{1}_{m+1},0,0,\ldots)$. Then $\|a^m\|_\infty = 1$ and $\|L(a^m)\|_1 = H_m$, then $m$th harmonic sum. Hence $\|L\|_{X_\infty\to X_1} \ge H_m$. Since $m$ was arbitrary and $(H_m)_{m\in \Bbb N}$ is a divergent sequence, then $\|L\|_{X_\infty\to X_1} = \infty$.