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Riesz-Representation Theorem for $L^p$ spaces says the following

Let $p \in [1,\infty]$, let $(X,\mu)$ be a measure space, let $T \in (L^{p}(X,\mu) )$ meaning that

$T \; : \; L^p(X,\mu) \to \mathbb{R}$

$T(a u + b v) = aT(u) + bT(v) \;\; \forall a,b \in \mathbb{R} \;,\, \forall u,v \in L^p(X,\mu)$

$|T(u)| \leq C||u||_{L^p(X,\mu)} \; \forall u \in L^p(X,\mu)$

Then if $p \in (1,\infty)$, or if $p = 1$ and $(X,\mu)$ is $\sigma$-finite then there exists a $g \in L^q(X,\mu)$ where $q \geq 1$ is s.t. $1/q + 1/p = 1$ such that

$T(u) = \int_X{ g \cdot u d\mu} \;\;\; \forall u \in L^p(X)$

if $p= \infty$ then $g$ may not exists, an example is $X = [-1,1]$, $\mu = \text{Lebesgue - measure}$, $T = \delta \;\;$ ($T(u) = u(0)$)

My question regards the case where $p = \infty$, in particolare what I'm asking is if the following statement is true

Statement :

Let $(X,\mu)$ be a $\sigma$-finite measure space

Let $T \; : \; L^\infty(X,\mu) \to \mathbb{R}$ be a continuos functional

Let $h \in L^1(X,\mu)$ be such that $h \geq 0$ and

$|T(u)| \leq \int_X{h |u| d\mu } \;\; \forall u \in L^\infty(X,\mu)$

Does there exist a function $g \in L^1(X,\mu)$ such that

$T(u) = \int_X{g u d\mu} \;\; \forall u \in L^\infty(X,\mu)$

How can I prove or disprove the statement? If the statement isn't true, is it true in the case where $X = \mathbb{R}^N$ and $\mu$ is the Lebesgue measure?

Paul
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1 Answers1

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It is true in general. By the reference provided in the comment above by copper.hat we can write $Tu=\int u d\nu$ for some finitely additive measure $\nu$.

Now let $d\lambda =hd\mu$. Then $\lambda$ is a finite positve measure and $|\int u d\nu|=|Tu| \leq \int |u|d\lambda$. Taking $u$ to be characteristic function of a set $A$ we get $|\nu (A)| \leq \lambda (A)$. This forces $\nu$ to to countably additive and we also get $\nu << \lambda$. Now a simple application of Radon Nikodym Theorem finishes the proof.

geetha290krm
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