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.

About this weblog

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”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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The Unapologetic Mathematician

Mathematics for the interested outsider