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$M$ 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!

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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
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. $\hat T(f)\geq0$.

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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

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