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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.

Watson
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Bombyx mori
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    See [this](http://math.stackexchange.com/questions/61458/if-f-is-measurable-and-fg-is-in-l1-for-all-g-in-lq-must-f-in-lp) post. – David Mitra Jan 02 '13 at 23:02
  • I am curious where the proof broke for the $p=1,\infty$ when it is not $\sigma$-finite. – Bombyx mori Jan 02 '13 at 23:16
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    For $p=1,\infty$ necessary and sufficient conditions and a counterexample when they fail are given [here](http://math.stackexchange.com/q/150547/). – Martin Jan 02 '13 at 23:27
  • I see. Now everything is pieced together. Thanks! – Bombyx mori Jan 03 '13 at 00:14
  • @DavidMitra: I am wondering what can I do to prove this without $\sigma$-finite assumption in $1

    – Bombyx mori Jan 03 '13 at 00:51
  • I see. I need to use the one for Hilbert spaces. – Bombyx mori Jan 03 '13 at 01:25
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    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

    – David Mitra Jan 03 '13 at 01:51
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    @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 03 '13 at 02:00
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    @DavidMitra: A look on the definition of uniformly convex space looked very familiar; I might have proved this before. So maybe this is Rudin's strategy. Thanks. – Bombyx mori Jan 03 '13 at 02:03
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    @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 03 '13 at 07:13
  • @Martin: Thank you! – Bombyx mori Jan 03 '13 at 07:21

1 Answers1

6

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.

Martin
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  • Give me some time to think about your solution carefully. – Bombyx mori Jan 03 '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 03 '13 at 20:18
  • Any progress? Do you need more details? – Martin Jan 08 '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 09 '13 at 06:26
  • Alright, thanks. Have fun with differentiation then :-) – Martin Jan 09 '13 at 07:07
  • Sorry for being slow :( – Bombyx mori Jan 09 '13 at 07:22
  • No worries, there's no hurry! No need to feel bad at all. I only wanted to ask about the state of affairs and I didn't intend to pressure you to do anything. Go at your own pace and do whatever you are up to. – Martin Jan 09 '13 at 07:25
  • I roughly finished differentiation, though still left a lot of problems unanswered. I will spend sometime today to think about your answer carefully. – Bombyx mori Jan 11 '13 at 10:13
  • This is pretty! I really enjoy it. – Bombyx mori Jan 11 '13 at 12:37
  • Thanks! As a PS: You can see here what breaks down in the case $p=1$: it is still true that we can find such functions $f_n$ and $E$, but for the norm of $\varphi$ we only have $$\lVert \varphi \rVert = \max\{\lVert \varphi|_{L^1(E)}\rVert, \lVert \varphi|_{L^1(X \setminus E)}\rVert\}.$$ This is insufficient to conclude $\varphi|_{L^1(X \setminus E)} = 0$. We can represent the restriction $\varphi|_{L^1(E)}$ by an $L^\infty(E)$-function, but on $X \setminus E$ there is still the possibility of non-vanishing. With this approach we can try to decompose $X$ into disjoint sets ... – Martin Jan 11 '13 at 13:00
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    ... 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
  • Let me think about your comments carefully; I think I learned a lot today. – Bombyx mori Jan 11 '13 at 13:07
  • I have one thing to ask; how do you get $|\phi|=(|(\phi)_{E}|^{q}+|(\phi_{X/E}|^{q}|)^{1/q}$? You cited duality, but I do not see how duality in $\sigma$-finite case give me this identity. Let me think about it. – Bombyx mori Jan 11 '13 at 13:15
  • This is an abstract fact: if a Banach space $X$ can be written as $X = Y \mathbin{\oplus_p} Z$ then its dual space $X^\ast$ can be written as $X^\ast = Y^\ast \mathbin{\oplus_q} Z^\ast$. This generalizes more or less directly from the duality between the two-dimensional spaces $(\mathbb{R}^2, \lVert \cdot \rVert_p)$ and $(\mathbb{R}^2, \lVert \cdot\rVert_q)$. – Martin Jan 11 '13 at 13:19
  • I am sorry for being slow. Assume $|f|_{p}=(|f|_{E}^{p}+|f|_{X/E}^{p})^{1/p}$ and $|\phi|_{q}=\max |\phi(f)|/|f|_{p}$, then the division is complicated. Since $|f|_{X/E}^{p}=0$, this become $|\phi(f)|/)|f_{E}|_{p}=|\phi|_{E}<\infty$. But I guess that's not what you intended. – Bombyx mori Jan 11 '13 at 13:35
  • Sorry I need to sleep; but I will read your answer carefully when I woke up. Thank you. – Bombyx mori Jan 11 '13 at 13:35
  • See also [here](http://math.stackexchange.com/q/236881) for a detailed answer. – Martin Jan 11 '13 at 13:44
  • Thanks! I feel learned a lot. – Bombyx mori Jan 12 '13 at 05:48