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Let $(X,A,\mu)$ be a measurable space. $1<p,q<\infty$, $\frac{1}{p}+\frac{1}{q}=1$. $Z\subset L^p$ a subspace of $L^p$ .Let $\phi: Z\to C$ a linear functional. Assume that there's $m<\infty$ such that for all $f\in Z$ such that $||f||_p<1$ : $|\phi(f)|\leq m$. Show that there's $g_0\in L^q$ such as $||g_0||\leq m$ and $\phi(f)=\int_X fg_0 d\mu$ for all $f\in Z$.

I tried to use the dual theorem. What i did: Let $g_0\in L^q$ (under the given conditions) Define $\phi: Z\to C$ as:

For $f\in L^p$ , $\phi(f)=\int_X fg_0 d\mu$. It is clear that $\phi$ is linear sincd the integral is linear.(and $\phi \in (L^p)^*$).

  1. $||\phi(f)||_C=|\int_X fg_0 d\mu|\leq \int_X |fg_0|d\mu=||fg_0||_1\leq ||f||_p||g_0||_q$. Here we used Holder inequality.

Definition of dual norm gives:

  1. $||\phi(f)||=sup |\phi(f)|_{||f||\leq 1}\leq m$

So in 1 we get by using 2 that:

$||\phi(f)||_C=|\int_X fg d\mu|\leq \int_X |fg|d\mu=||fg||_1\leq ||f||_p||g||_q \leq 1*||g_0||_q \leq m$.

Unfortunately, this does not show that there is $g_0$ that satisfy the given conditions.

Lam18373
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    Your proof doesn't show that given any $\phi$ you can find such $g_{0} \in L_{q}$, only that any function $g_{0} \in L_{q}$ defines such a functional. – Tom Ariel Dec 31 '20 at 18:27
  • Hi @Tom Ariel can you explain how to show this correctly? – Lam18373 Jan 01 '21 at 09:12
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    This sounds like a Hahn-Banach theorem problem to me, but maybe I'm misinterpreting. – Cameron L. Williams Jan 01 '21 at 15:11
  • @Cameron Williams , me too but how to connect can this be hepful/useful? – Lam18373 Jan 01 '21 at 15:24
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    Your conditions on $\phi$ are that it is a continuous linear map on $Z$. Hahn Banach gives you an extension $\widetilde \phi: L^p(X)\to\Bbb C$, ie an element of the dual of $L^p(X)$. If $p\neq\infty$ then this is simply $L^q(X)$. The way $L^q(X)$ acts on $L^p(X)$ is by integrating. – s.harp Jan 01 '21 at 16:10
  • @s.harp , that is connected to the dual theory right? Can you please add an answer to this? It would be helpful to understand it better:) since i am not vetting why there exists such $g_0$ – Lam18373 Jan 01 '21 at 16:16

1 Answers1

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If $\phi$ is a linear functional $Z\to\Bbb C$ so that $|\phi(f)|≤m$ whenever $\|f\|<1$ then $\phi$ is necessarily continuous and the operator norm of $\phi$ is $≤m$. You may then apply Hahn-Banach to extend $\phi$ to a linear functional $\tilde \phi: L^p(X)\to\Bbb C$, the Hahn Banach extension will also satisfy $\|\tilde \phi\|≤m$.

Now if $p\in [1,\infty)$ then the dual of $L^p(X)$ is equal to $L^q(X)$ where $q$ is so that $\frac1p+\frac1q=1$. Hence you may identify $\tilde\phi$ with an element $g_0\in L^q(X)$. The dual action of $L^q(X)$ on $L^p(X)$ is given by integrating, for $f\in L^p(X)$ and $g\in L^q(X)$ you have: $$g(f) := \int_X g(x)\cdot f(x)\ d\mu(x)$$ Hence you find for all $f\in L^p(X)$ that $$\tilde \phi(f) = \int_X g_0(x)\cdot f(x)\ d\mu(x)$$ in particular since $\tilde\phi\lvert_{Z}=\phi$ you have that for all $f\in Z$ you've got: $$\phi(f) = \int_X g_0(x)\cdot f(x)\ d\mu(x)$$ One more statement about the duality $L^p(X)^*\cong L^q(X)$ needs to be referenced, namely that the norms $$\|g\|_{op}:= \sup_{f\in L^p(X), \ \|f\|≤1} |g(f)|\quad \text{ and }\quad\|g\|_q := \sqrt[q]{\int_X |g(x)|^q\ d\mu(x)}$$ are equal. Hence $\|g_0\|_q =\|\tilde\phi\|_{op} ≤ m$.

The statements about the duality $L^p(X)^* = L^q(X)$ can be found eg on wikipedia or also on this site (here is something that came up in a search: Duality of $L^p$ spaces)

s.harp
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