Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$M$ is the subspace of $L_p[a,b]$ that $\forall f\in L_p[a,b]$ $\exists g\in M$ with $f(t)\leq g(t)$ almost everywhere.

$T:M\rightarrow \mathbb{R}$ $\quad$$T(f)\geq0$ whennever $f(t)\geq0$ a.e

Question: Show that We can extend $T$ to all $L_p[a,b]$, in other words $$\exists\quad \hat{T}:L_p[a,b]\rightarrow\mathbb{R}$$ that $\hat{T}(f)\geq0 $ whennever $f(t)\geq0$ a.e. and $\hat{T}|_M=T$.

And the hint is: use as sub-linear $p(f)=\inf\{T(g) \quad |\quad g\in M , f\leq g\}$

I was trying to use this version of Hahn Banach:

$X$ real vector space, $M$ a subspace, $p$ a sub-linear mapping and $T:M\rightarrow\mathbb{R}$ linear and $T(x)\leq p(x)$ $\forall x \in M$ Then $\exists$ an extension of $T$ to whole X. And $\hat{T}\leq p(x) \quad\forall x\in X$.

So to be able to apply this theorem we must check the following:

  1. Show $T$ is linear.
  2. $p$ is sub-linear, in other words, $p(f+g)\leq p(f)+ p(g)$ and $p(\alpha f)=\alpha p(f)\quad\forall \alpha\geq 0 f,g\in L_p[a,b]$
  3. Show $T(f)\leq p(f) \quad\forall f\in M$

How do we show $T$ is linear if we do not know how $T$ is defined? Is there some proposition that can tell us that $T$ will be linear?

I am very bad at maths, I think this version of Hahn Banach theorem will show us the answer of the problem, is this the correct version of H-B theorem ? Some hints for 2. and 3. will be helpful too.

Thank you very much!

share|improve this question
    
Your definition of $M$ at the beginning doesn't make sense to me. –  Martin Argerami Oct 31 '13 at 0:30
    
I will check the exercise. I did not understand the exercise so i am asking help thank you for the comment –  Maths Student Oct 31 '13 at 6:05
    
I think $M $ is a subspace is by hypothesis. stop and an extra information of $M$ is that for all $f$ in $X \exists$ a $g \in M $..... –  Maths Student Oct 31 '13 at 8:11
    
$X=L_p[a,b]$ i was thinking in the theorem, sorry –  Maths Student Oct 31 '13 at 8:40
    
May be the application of the theorem is not direct. I mean may we need an auxiliar function $F$, we define $F$ related with $T$, but we define $F$ as linear. Then we extend $F$ to $\hat{F}$ with the theorem but keeping the property of $T(f)\geq 0$ Is there someone want to help me, give me some hints please? –  Maths Student Oct 31 '13 at 11:49

1 Answer 1

up vote 1 down vote accepted

I will assume that the problem starts with "$M$ is a subspace..."

First $T$ has to be linear by definition. Otherwise no relation can be drawn to any subspace.

The fact that $p$ is sublinear: fix $f,g \in L_p$. For any $f',g'\in M$ such that $f\leq f'$ and $g\leq g'$, we have $$ p(f+g)\leq T(f'+g')=T(f')+T(g'). $$ So $p(f+g)-T(g')\leq T(f')$ for any $f'\in M$ such that $f\leq f'$. This shows that $$ p(f+g)-T(g')\leq p(f). $$ As $g'$ was arbitrary, we have $p(f+g)-p(f)\leq T(g')$ for any $g'\in M$ such that $g\leq g'$. So $$ p(f+g)-p(f)\leq p(g), $$ which shows the subadditivity.

To apply the Hahn-Banach Theorem you quote, all that is missing is that $T(f)\leq p(f)$ for all $f\in M$. But this occurs by definition: since $f\in M$, it is not hard to see that $p(f)=T(f)$: for any $g\in M$ with $f\leq g$, $T(f)\leq T(g)$ (because $T(g-f)\geq0$ and $T$ is linear).

So the Hahn-Banach Theorem gives us the existence of $\hat T$ with $\hat T\leq p$ and $\hat T|_M=T$. All that remains to see is that $\hat T\geq0$.

So fix $f\geq0$. Then there exists $g\in M$ with $f\leq g$, and thus $$ \hat T(f)\leq p(f)\leq T(g)=\hat T(g). $$ Then $\hat T(g-f)=T(g)-\hat T(f)\geq0$. Now, since $g-f\leq g$ and $g\in M$, $$ \hat T(g-f)\leq p(g-f)\leq T(g). $$ Then $T(g)\leq T(g)+\hat T(f)$, i.e. $\hat T(f)\geq0$.

share|improve this answer
    
And then the big space $L_p[a,b]$ did not do any intervention? I am little confused because in maths, normally, we use all hypothesis no? Thank you! –  Maths Student Nov 3 '13 at 16:50
    
Almost the last part, when you say "since $g-f\geq g$ " Why? We have $f\geq0$ a.e and $f\leq g$ a.e, but I can see why $g-f\geq g$ –  Maths Student Nov 3 '13 at 17:05
    
I think you wanted to say $g-f\leq g$, you just typed it wrong? –  Maths Student Nov 3 '13 at 17:19
    
Yes I did. I just edited it, thank you. –  Martin Argerami Nov 3 '13 at 17:32
    
Regarding your first question, you are not posing boundedness for $T$ (nor closedness for $M$?) so there is no real relation to the $p$-norm, as far as I can see. –  Martin Argerami Nov 3 '13 at 17:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.