3
$\begingroup$

This question was originally posted on MSE (https://math.stackexchange.com/q/3796602/793374), but nobody has found a correct answer in about two weeks, so I decided to repost it here:

In lemma 4.30 of Folland's "A Course in Abstract Harmonic Analysis" (Second Edition) one needs to show the square-integrability of the function $f$ defined below and I don't understand how Folland deduces it from the inequality below.

enter image description here

enter image description here

For context, $G$ is a locally compact abelian Hausdorff group and $dx$ is a Haar measure on $G$ (note that Folland defines Radon measures to be outer regular and inner regular on open sets). Furthermore $f$ is continuous, bounded and a linear combination of functions of positive type (this is encoded in the notation $f \in \mathcal{B}(G)$). Here is what i have tried so far:

With Plancherel's theorem we see (as in Folland's proof) that $$(L^1(G) \cap L^2(G), \|\cdot\|_2) \to \mathbb{C}, \ k \mapsto \int_G f(x) \cdot k(x) \, dx$$ defines a bounded linear functional which extends to a bounded linear functional $F \in L^2(G)^*$ by the BLT theorem (I removed the complex conjugation for linearity; this should not make a difference in the argumentation).

Now Riesz's theorem yields an $r \in \mathcal{L}^2(G)$ such that $F$ is given by integration against $r$, i.e. $$F(k) = \int_G r(x) \cdot k(x) \, dx \ \text{ for all } k \in L^2(G).$$ In particular we have $$\int_G f(x) \cdot k(x) \, dx = \int_G r(x) \cdot k(x) \, dx \ \text{ for all } k \in L^1(G) \cap L^2(G).$$

With this we can show that the set $N := \{x \in G: r(x) \neq f(x)\}$ is locally null with respect to the Haar measure $dx$ since for any Borel set $A \subseteq N$ with finite Haar measure we can set $$k(x) := 1_A(x) \cdot \frac{|f(x) - r(x)|}{(f(x) - r(x)) + 1_{G \setminus N}(x)}$$ to obtain a function $k \in L^1(G) \cap L^2(G)$, so $$0 = \int_G (f(x) - r(x)) \cdot k(x) \, dx = \int_A |f(x) - r(x)| \, dx,$$ i.e. $A \cap N = A$ has Haar measure $0$.

To conclude $f \in L^2(G)$ we now need to show that $N$ has Haar measure $0$ and this problem can be reduced further: The set $$R := \{x \in G: r(x) \neq 0\} = \bigcup_{n \in \mathbb{N}} \{x \in G: |r(x)| \geq \tfrac{1}{n}\}$$ is $\sigma$-finite since $r \in \mathcal{L}^2(G)$, so $R \cap N$ is again $\sigma$-finite and locally null. Hence $R \cap N$ has Haar measure $0$ and we only need to show that the set $$M := (G \setminus R) \cap N = \{x \in G: r(x) = 0 \neq f(x)\}$$ has Haar measure $0$.

$\endgroup$
1
  • 1
    $\begingroup$ If I recall correctly, Folland adopts a style/approach where most of the arguments are stated in a way that works for groups with a sigma-finite Haar measure, and he has some comments earlier in the book on how one can reduce to this case or adapt the arguments. Is this "hand-waving" the part that concerns you? $\endgroup$
    – Yemon Choi
    Sep 1 '20 at 0:37
4
$\begingroup$

In the following attempt, I am shameless using the fact that $f$ is continuous and bounded. My philosophy on $L^p$-spaces is shaped heavily by Banach space theory rather than measure theory, and most of my experience is with $\sigma$-finite measure spaces, so I apologize if I have missed some subtleties or conversely if I have belaboured some easy points.


Pick a compact $K\subset G$ and set $k=1_K\cdot f$; this certainly belongs to every $L^p(G)$ since $f$ is continuous and bounded, and since Haar measure is finite on compact sets.

Then, using the inequality that you quote from Folland, $$ \int_K f\overline{f} \,dx \leq {\Vert \phi \Vert}_2 \left( \int_K |f|^2 \right)^{1/2}$$ so that $$ \int_K |f(x)|^2\,dx \leq {\Vert\phi\Vert}_2^2 $$ (I think this is what someone was suggesting on MSE.) Now we are done provided we can justify the following claim.

Claim: Let $h\geq 0$ be a non-negative continuous bounded function on $G$, and let $\mu$ be a Radon measure on $G$. Then $$ \int_G h\,d\mu = \sup_K \int_K h\,d\mu $$ where the supremum is over all compact $K\subseteq G$.

(Note that I am not assuming that $G$ is $\sigma$-compact.)

Proof of claim. If the RHS is infinite there is nothing to prove; so we may assume it is finite, and denote this supremum by $C$. Clearly the LHS is $\geq C$ so we only need to establish the converse inequality.

Given $r \in (0,1)$, let $E_r= \{ x\in G \colon h(x) > r \}$. This is open, so by inner regularity of $\mu$ on open sets, there is an increasing sequence of compact subsets $K_1 \subseteq K_2 \subseteq \dots \subseteq E$ with $\mu(K_n) \nearrow \mu(E_r)$. But then, using our assumption, $$ r\mu(K_n) \leq \int_{K_n} h\,d\mu \leq C \qquad\hbox{for all $n$} $$ and so we have $\mu(E_r) \leq C/r<\infty$.

Since $E_r$ has finite measure and $\sup_n\mu(K_n)=\mu(E_r)$, the set $E_r \setminus \bigcup_{n\geq 1} K_n$ has measure zero and we have $$ \int_{E_r} h\,d\mu = \lim_n \int_{K_n} h\,d\mu \leq C. $$ By taking $r\searrow 0$ along some decreasing sequence in$(0,1)$, it follows that $\int h\,d\mu \leq C$, as required.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer! One more question: Is boundedness of $h$ actually required for the claim you prove? I don't see it used anywhere (since the two integral limits should follow from monotone convergence), but I might be missing something. $\endgroup$ Sep 1 '20 at 12:23
  • 1
    $\begingroup$ I think you are right: I was making the proof up as I went along, and in an earlier version I wanted to play safe by assuming h is bounded. It also seems that lower semi-continuity of h is sufficient $\endgroup$
    – Yemon Choi
    Sep 1 '20 at 14:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.