The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

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