# Square-integrability in lemma 4.30 of Folland's "A Course in Abstract Harmonic Analysis"

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.

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 $$0$$.

• 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? Sep 1 '20 at 0:37

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.

• 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. Sep 1 '20 at 12:23
• 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 Sep 1 '20 at 14:49