Some Continuous Duals

I really wish I could just say $L^p$ in post titles.

Anyway, I want to investigate the continuous dual of $L^p(\mu)$ for $1\leq p<\infty$ . That is, we’re excluding the case where either $p$ (but not its Hölder conjugate $q$ ) is infinite. And I say that when $(X,\mathcal{S},\mu)$ is $\sigma$ -finite, the space $L^p(\mu)'$ of bounded linear functionals on $L^p(\mu)$ is isomorphic to $L^q(\mu)$ .

First, I’m going to define a linear map $\kappa_p:L^q\to\left(L^p\right)'$ . Given a function $g\in L^q$ , let $\kappa_p(g)$ be the linear functional defined for any $f\in L^p$ by

$\displaystyle\left[\kappa_p(g)\right](f)=\int fg\,d\mu$

It’s clear from the linearity of multiplication and of the integral itself, that this is a linear functional on $L^p$ . Hölder’s inequality itself shows us that not only does the integral on the right exist, but

$\displaystyle\lvert\left[\kappa_p(g)\right](f)\rvert\leq\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$

That is, $\kappa_p(g)$ is a bounded linear functional, and the operator norm $\lVert\kappa_p(g)\rVert_\text{op}$ is at most the $L^q$ norm of $g$ . The extremal case of Hölder’s inequality shows that there is some $f$ for which this is an equality, and thus we conclude that $\lVert\kappa_p(g)\rVert_\text{op}=\lVert g\rVert_q$ . That is, $\kappa_p:L^q\to\left(L^p\right)'$ is an isometry of normed vector spaces. Such a mapping has to be an injection, because if $\kappa_p(g)=0$ then $0=\lVert\kappa_p(g)\rVert_\text{op}=\lVert g\rVert_q$ , which implies that $g=0$ .

Now I say that $\kappa_p$ is also a surjection. That is, any bounded linear functional $\Lambda:L^p\to\mathbb{R}$ is of the form $\kappa_p(g)$ for some $g\in L^q$ . Indeed, if $\Lambda=0$ then we can just pick $g=0$ as a preimage. Thus we may assume that $\Lambda$ is a nonzero bounded linear functional on $L^p$ , and $\lVert\Lambda\rVert_\text{op}>0$ . We first deal with the case of a totally finite measure space.

In this case, we define a set function on measurable sets by $\lambda(E)=\Lambda(\chi_E)$ . It’s straightforward to see that $\lambda$ is additive. To prove countable additivity, suppose that $E$ is the countable disjoint union of a sequence $\{E_n\}$ . If we write $A_k$ for the union of $E_1$ through $E_k$ , we find that

$\displaystyle\lVert\chi_E-\chi_{A_k}\rVert_p=\left(\mu(E\setminus A_k)\right)^\frac{1}{p}\to0$

Since $\Lambda$ is continuous, we conclude that $\lambda(A_k)\to\lambda(E)$ , and thus that $\lambda$ is a (signed) measure. It should also be clear that $\mu(E)=0$ implies $\lambda(E)=0$ , and so $\lambda\ll\mu$ . The Radon-Nikodym theorem now tells us that there exists an integrable function $g$ so that

$\displaystyle\Lambda(\chi_E)=\lambda(E)=\int\limits_Eg\,d\mu=\int\chi_Eg\,d\mu$

Linearity tells us that

$\displaystyle\Lambda(f)=\int fg\,d\mu$

for simple functions $f$ , and also for every $f\in L^\infty(\mu)$ , since each such function is the uniform limit of simple functions. We want to show that $g\in L^q$ .

If $p=1$ , then we must show that $g$ is essentially bounded. In this case, we find

$\displaystyle\left\lvert\int\limits_Eg\,d\mu\right\rvert\leq\lVert\Lambda\rVert_\text{op}\,\lVert\chi_E\rVert_1=\lVert\Lambda\rVert_\text{op}\mu(E)$

for every measurable $E$ , from which we conclude that $\lvert g(x)\rvert\leq\lVert\Lambda\rVert_\text{op}$ a.e., or else we could find some set on which this inequality was violated. Thus $\lVert g\rVert_\infty\leq\lVert\Lambda\rVert_\text{op}$ .

For other $p$ , we can find a measurable $\alpha$ with $\lvert\alpha\rvert=1$ so that $\alpha g=\lvert g\rvert$ . Setting $E_n=\{x\in X\vert n\geq\lvert g(x)\rvert\}$ and defining $f=\chi_{E_n}\lvert g\rvert^{q-1}\alpha$ , we find that $\lvert f\rvert^p=\lvert g\rvert^q$ on $E_n$ , $f\in L^\infty$ , and so

$\displaystyle\int\limits_{E_n}\lvert g\rvert^q\,d\mu=\int\limits_Xfg\,d\mu=\Lambda(f)\leq\lVert\Lambda\rVert_\text{op}=\lVert\Lambda\rVert_\text{op}\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^\frac{1}{p}$

We thus find

$\displaystyle\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^\frac{1}{q}=\left(\int\limits_{E_n}\lvert g\rvert^q\,d\mu\right)^{1-\frac{1}{p}}\leq\lVert\Lambda\rVert_\text{op}$

and thus

$\displaystyle\int\limits_X\chi_{E_n}\lvert g\rvert^q\,d\mu\leq\lVert\Lambda\rVert_\text{op}^q$

Applying the monotone convergence theorem as $n\to\infty$ we find that $\lVert g\rVert_q\leq\lVert\Lambda\rVert_\text{op}$ .

Thus in either case we’ve found a $g\in L^q$ so that $\Lambda=\kappa_p(g)$ .

In the $\sigma$ -finite case, we can write $X$ as the countable disjoint union of sets $X_i$ with $\mu(X_i)<\infty$ . We let $Y_k$ be the union of the first $k$ of these sets. We note that $\lVert\chi_E f\rVert_p\leq\lVert f\rVert_p$ for every measurable set $E$ , so $f\mapsto\Lambda(\chi_Ef)$ is a linear functional on $L^p$ of norm at most $\lVert\Lambda\rVert_\text{op}$ . The finite case above shows us that there are functions $g_i$ on $X_i$ so that

$\displaystyle\Lambda(\chi_{X_i}f)=\int\limits_{X_i}fg_i\,d\mu$ .

We can define $g_i(x)=0$ if $x\notin X_i$ , and let $g$ be the sum of all these $g_i$ . We see that

$\displaystyle\Lambda(\chi_{Y_k}f)=\int\limits_{Y_k}f(g_1+\dots+g_k)\,d\mu$

for every $f\in L^p$ , and since $\mu(Y_k)<\infty$ we find that $\lVert g_1+\dots+g_k\rVert_q\leq\lVert\Lambda\rVert_\text{op}$ . Then Fatou’s lemma shows us that $\lVert g\rVert_q\leq\lVert\Lambda\rVert_\text{op}$ . Thus the $\sigma$ -finite case is true as well.

One case in particular is especially worthy of note: since $2$ is Hölder-coonjugate to itself, we find that $L^2$ is isomorphic to its own continuous dual space in the same way that a finite-dimensional inner-product space is isomorphic to its own dual space.

September 3, 2010 - Posted by John Armstrong | Analysis, Functional Analysis, Measure Theory

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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