# How should I prove the duality?

Rudin asked (Real Complex Analysis, First edition, Chapter 6, Problem 4):

Suppose $1\le p\le \infty$, and $q$ is the exponent conjugate to $p$. Suppose $u$ is a $\sigma$-finite measure and $g$ is a measurable function such that $fg\in L^{1}(\mu)$ for every $f\in L^{p}(\mu)$. Prove that then $g\in L^{q}(\mu)$.

I am being troubled with the fact that $|g|_{q}$ might be unbounded if we select wierd enough $f$. Holder inequality only gives us $$|fg|_{1}\le |f|_{p}|g|_{q}\leftrightarrow |g|_{q}\ge \frac{|fg|_{1}}{|f|_{p}}$$All the constructions I know proving $g\in L^{q}$ starts by assuming $f\rightarrow fg$ is a bounded linear operator，so I cannot use circular reasoning at here.

It suffice to prove the statement for finite measure spaces and simple functions. So emulating Rudin we can assume $|g|=\alpha g$, where $|\alpha|=1$ and $\alpha$ is measurable. Let $E_{n}=x:|g(x)|\le n$ and let $f=\chi_{E_{n}}\alpha g^{q-1}$. Then we have $$\int_{E_{n}}|g|^{q}d\mu=\int_{X}|fg|d\mu\le K_{n}$$ for some $K_{n}<\infty$. But this constant obviously shift with the $n$ I choose, hence probably does not have a finite upper bound (for example $K_{n}=n$). And I got stuck.

In problem 6, Rudin now ask:

Suppose $1<p<\infty$, and prove that $L^{q}(\mu)$ is the dual space of $L^{p}(\mu)$ even if $\mu$ is not $\sigma$-finite.

I keep thinking about it but do not know what is the best way to prove it.

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See this post. –  David Mitra Jan 2 '13 at 23:02
For $p=1,\infty$ necessary and sufficient conditions and a counterexample when they fail are given here. –  Martin Jan 2 '13 at 23:27
I don't think the case where the measure is not $\sigma$-finite is so easy. The proofs I've seen use the fact that for $1<p<\infty$, $L_p$ is uniformly convex. One proof using this approach can be found on page 330 of these notes (which uses Clarkson's inequality to show $L_p$ is uniformly convex for $2\le p<\infty$. Another proof (also using Clarkson's inequality) can be found in Hewitt and Stromberg's Real and Abstract Analysis. –  David Mitra Jan 3 '13 at 1:51
@DavidMitra: Yes, that's why I asked. I do not think Rudin will assume I know uniform convex or Clarkson's inequality, and he did not introduce Riesz representaion theorem for Hilbert space in that chapter. So he must have imagined some different proof. –  Bombyx mori Jan 3 '13 at 2:00
@David: You can reduce to the $\sigma$-finite case. That the map $L^q \to (L^p)^\ast$ is isometric is easy, so surjectivity is at stake. The idea is that a functional $\varphi$ on $L^p$ must be "supported" on a $\sigma$-finite set: if $f_n \in L_p$ are normalized functions such that $\varphi(f_n) \to \lVert \varphi\rVert$ then the set $E_n = \{x : \sum_{i=1}^n |f_i(x)|^p \geq 2^{-n}\}$ has finite measure and one shows that $\varphi$ cannot charge functions with support entirely outside $E = \bigcup E_n$. Now localize to $E$ and apply the argument from the $\sigma$-finite case. –  Martin Jan 3 '13 at 7:13

This answer is about the last part on how to extend the duality beyond $\sigma$-finite measures since the first part is taken care of in If $f$ is measurable and $fg$ is in $L^1$ for all $g \in L^q$, must $f \in L^p$? mentioned by David Mitra.

The answer to "what is the best way to prove that $L^{q}(\mu)$ is the dual space of $L^{p}(\mu)$ even if $\mu$ is not $\sigma$-finite" is probably a matter of taste. I think the proof using uniform convexity and the Milman-Pettis theorem suggested by David Mitra is nice. The usual expositions apply Clarkson's inequalities or Hanner's inequalities. The latter are easier to prove than the former, but they leave me with the feeling of coming out of the blue, see also the discussion on Mathoverflow.

An illuminating discussion of the approach to duality via uniform convexity of the $L^p$-spaces is given by Harald Hanche-Olsen, On the uniform convexity of $L^p$, Proc. Amer. Math. Soc. 134 (2006), 2359-2362. The paper ends with an outline of the proof you're asking about. Alternatively, you can consult the references provided by David Mitra in the comments.

However, I doubt that the above-mentioned proofs are what Rudin had in mind in this exercise. After all, we want to prove that $L^q$ is dual to $L^p$ for $\frac{1}{p} + \frac{1}{q} = 1$ by using the methods he provides in said chapter and by reducing to the $\sigma$-finite case.

It is not hard to prove that the natural map $L^q \to (L^p)^\ast$ given by sending $g \in L^q$ to $\varphi_g(f) = \int f g$ is isometric: this uses Hölder's inequality and one verifies that $f = \lVert g\rVert_{p}^{1-p} \frac{\lvert g\rvert}{g}$ is an $L^p$ function of norm $\lVert f \rVert_q = 1$ such that $\varphi_g(f) = \lVert g\rVert_q$.

It remains to prove that every $\varphi \in (L^p)^\ast$ is of the form $\varphi_g$ for some $g \in L^q$.

Observe that a function $g \in L^q$ must vanish outside of a set of $\sigma$-finite measure (since $\lvert g\rvert^q$ is integrable). So, given a continuous functional $\varphi \colon L^p \to \mathbb{R}$, we must somehow be able to prove that it is "supported" on a set of $\sigma$-finite measure. By this I mean that we must be able to find a $\sigma$-finite set $E$ such that $f|_E = 0$ implies $\varphi(f) = 0$. If we have that, the known duality applies.

Let $f_n \in L^p$ be functions of norm $\lVert f_n\rVert_p = 1$ such that $\varphi(f_n) \to \lVert \varphi \rVert$. The intuition is that $\lVert \varphi\rVert-\varepsilon \lt \varphi(f_n) \leq \lVert \varphi\rVert$ means that $\varphi$ must almost be zero on functions whose support is disjoint from the one of $f_n$.

Collect all the supports of the $f_n$'s together into one set $E$, using a small trick to show that it is $\sigma$-finite: let $E_n = \{x \in X : \sum_{i=1}^n |f_i(x)|^p \geq 2^{-n}\}$. We have $\mu(E_n) \lt \infty$ and $E = \bigcup_{n=1}^\infty E_n$ is a set of $\sigma$-finite measure containing the supports of all the $f_n$'s. If we manage to prove that $\varphi(h) = 0$ for all $h \in L^p$ such that $h|_E = 0$ we're done: $E$ is $\sigma$-finite and thus we find a function $g$ in $L^q(E)$ representing $\varphi$ (extend $g$ by zero on $X \setminus E$).

Notice that $L^p(X) = L^p(E) \mathbin{\oplus_p} L^p(X \setminus E)$, by which I mean that each $f \in L^p(X)$ can be uniquely written as $f = f|_E + f|_{X \setminus E}$ and we have the identity $$\lVert f \rVert_{p} = \left(\lVert f|_E \rVert_p^{p} + \lVert f|_{X \setminus E}\rVert_{p}^p\right)^{1/p}$$ for the norm. By duality and the assumption $1 \lt p \lt \infty$ we get $$\lVert \varphi\rVert = \left(\lVert\varphi|_{L^p(E)}\rVert^{q} + \lVert \varphi|_{L^{q}(X\setminus E)}\rVert^q\right)^{1/q},$$ see $(X \oplus_p Y)^*$ isometric to $(X^*\oplus_q Y^*)$ for details. By the choice of $E$ we have $\lVert\varphi\rVert = \lVert \varphi|_{L^p(E)}\rVert$ whence $\varphi|_{L^p(X \setminus E)} = 0$, as claimed earlier.

To repeat, $E$ is a $\sigma$-finite set, we can find $g \in L^q(E)$ such that $\varphi_g = \varphi|_{L^p(E)}$ and by setting $g = 0$ on $X \setminus E$ we get a function in $L^q(X)$ representing $\varphi$. Thus, the map $L^q(\mu) \to L^p(\mu)^\ast$ is surjective.

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Give me some time to think about your solution carefully. –  Bombyx mori Jan 3 '13 at 19:49
Sure, take all the time you need. I deliberately left some details for you to fill in. Nothing should be too hard, please don't hesitate to ask for further explanations. –  Martin Jan 3 '13 at 20:18
Any progress? Do you need more details? –  Martin Jan 8 '13 at 12:12
Sorry for replying so late. I am reading Rudin's Chapter on Differentiability theorems. Once I finished that I will read your answer carefully. –  Bombyx mori Jan 9 '13 at 6:26
... of $\sigma$-finite measure on each of which we can find an $L^\infty$-function representing $\varphi$. Then we need a further condition guaranteeing that we can "patch together" these measurable functions on these $\sigma$-finite pieces to a measurable function on all of $X$. It turns out that many natural measure spaces do have this property and, moreover, that this property --- carefully phrased --- is not only sufficient, but also necessary to have the duality between $L^1$ and $L^\infty$. (references in the thread linked in my first comment to your question). –  Martin Jan 11 '13 at 13:03