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Consider the following fragment from Folland's book "A course in abstract harmonic analysis". Here $G$ is a locally compact Hausdorff group and the $L^p$-spaces are considered w.r.t. a left Haar measure on $G$.

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The proof of proposition 2.40 applies Minkowski's integral inequality, however the proofs of Minkowski's integral inequality I know make crucial use of the fact that the measure spaces involved are $\sigma$-finite. However, $G$ with the left Haar measure is $\sigma$-finite if and only if $G$ is $\sigma$-compact, which is not an assumption here.

How can I solve this technicality? Perhaps I can make use of the fact that if $f \in L^1(G)$, then $f$ vanishes outside a $\sigma$-finite set and similarly for $g \in L^p(G)$ with $p < \infty$. Can someone fill in the details? And shouldn't the case $p=\infty$ be dealt with on its own?

Andromeda
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    Typically, the group $G$ is assumed to be locally compact Hausdorff so that the Haar measure is indeed a Radon measure. Many important topological groups are $\sigma$-finite w.r.t Haar measure. For more general topological groups, notice that the integrability of $f$ and $g$ (or rather $|f|$ and $|g|^p$, imply that the carriers of $f$ and $g$ (i.e. the sets ${x\in G: |f(x)|>0}$ and ${x\in G: |g(x)|>0}$) are $\sigma$-finite. Thus, the carrier of $(x,y)\mapsto f(y)g(y^{-1}x)$ will also be $\sigma$-finite (w.r.t. the product Haar measure) – Mittens Apr 19 '22 at 14:07
  • @OliverDíaz Thanks! But doesn't this break down if $p=\infty$? – Andromeda Apr 19 '22 at 14:28
  • Folland's book on Abstract Harmonic Analysis, section 2.3: Interlude: Some Technicalities, p.p. 43-46 explains this and some other problems that may arise when dealing with topological groups that are not $\sigma$-compact. This groups contain a subgroup $H$ that is both open and closed and which is $\sigma$-compact. That will help a lot. Finally, notice that for $1\leq p<\infty$, $C_{00}(G)$ is dense in $L_p$; this $L_p$ functions vanish outside a $\sigma$-compact set. – Mittens Apr 19 '22 at 14:28

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This is to address only the case $p=\infty$.

Let $\lambda$ be a Haar measure (left invariant) on the locally compact Hausdorff topological group $G$. Let $f\in L_1(\lambda)$ and $g\in L_\infty(\lambda)$. Then $$f*g(x):=\int_G f(y)g(y^{-1}x)\lambda(dy)=\int_G f(xy)g(y^{-1})\,\lambda(dy)$$ If $\phi(y):=g(y^{-1})$, notice that $|\phi(y)|\leq\|g\|_{\infty}$ for $\lambda$-a.s $y\in G$ (the measure $\rho(A)=\lambda(A^{-1})$ preserves sets of measure zero, in fact $\rho(dy)=\triangle(y^{-1})\lambda(dy)$, where $\triangle$ is the modular function). From this, it follows that $$|f*g(x)|\leq\|g\|_\infty\int_G|f(xy)|\,\lambda(dy)=\|g\|_\infty\|f\|_1$$

Comment: With a little more effort one can probe that in fact $f*g\in\mathcal{C}(G)$.

Mittens
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  • Thanks again for the help. Perhaps one point of clarification (not immediately related to your answer): Should I interpret $L^\infty(G)$ in the theorem in the usual sense (i.e. as essentially bounded measurable functions with the usual $|\cdot|_\infty$-norm) or as defined by Folland on p50-51 (so locally measurable functions that are bounded locally a.e.)? Or doesn't it matter? Just trying to understand the technicalities in harmonic analysis! – Andromeda Apr 19 '22 at 15:58
  • @Andromeda: Those are all good questions. The essential $|f|\infty$ and locally essential $|f|^{\ell}\infty$ suprema are not the same. They do coincide when the underlying space is $\sigma$-finite (for example this posting. In the general setting you are dealing (general locally compact Hausdorff topological groups) the essential supremum is convenient as it preserves the duality $(L^1)^*=L_\infty$. Also, the carrier $\kappa$ of $f$ reduces the integral $\int_G$ to $\int_\kappa$ and then this are fie since $\kappa$ is $\sigma$-finite. – Mittens Apr 19 '22 at 16:21
  • Don't you mean locally essential supremum (... is convenient)? Also, I could not make sense of your last sentence (is there a typo?). – Andromeda Apr 19 '22 at 16:35
  • @Andromeda: It is convenient to work with $L_\infty$ with the local essential supremum so that duality $L_1$ and $L_\infty$ is preserved. As for the second part of my comment, what I tied to convey is that in the integral $\int_G f(y)g(y^{-1}x),\lambda(dy)$ can be expressed as $\int_{\kappa}f(y) g(y^{-1}x),\lambda(dy)$ where $K$ is $\sigma$ finite ($K={y\in G: |f(y)|>0}$ is $\sigma$-finite due to the integrability of $f$). – Mittens Apr 19 '22 at 17:07
  • Thanks! So by your last comment, it follows that it doesn't matter which definition of $L^\infty$ because we are essentially integrating over something $\sigma$-finite, right? – Andromeda Apr 19 '22 at 17:23
  • Thanks for all the help! – Andromeda Apr 19 '22 at 17:37
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    Your welcome! you did push my little grey cells to work with your last two postings. The thing is that once you sort of understand the nuances of Fubini's theorem, one tends to pass arguments about a..s. and so on as granted. Questioning such nuances forces one to go back to the guts of what Fubini's theorem says and some details about its proof. – Mittens Apr 19 '22 at 17:40
  • Sometimes I wish I wouldn't care about such technicalities, I would go a lot faster through these books! – Andromeda Apr 19 '22 at 18:05
  • Haha I won't be dealing with stochastic integration anytime soon I think. – Andromeda Apr 19 '22 at 18:39
  • I added another answer to fill in another detail that Folland left out. Or is this somehow obvious from Folland's proof? It appears to me when he writes $|f\star g|_p$ that he has already established that $f \star g$ makes sense! – Andromeda Apr 21 '22 at 08:33
  • Also, I disagree that it doesn't matter which definition of $L^\infty$ we use here. We should use Folland's definition of $L^\infty$ here: this gives a sharper bound and implies the result for the usual $L^\infty$ as well. – Andromeda Apr 21 '22 at 17:04
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This is to show that for $1\leq p<\infty$, one can use the generalized Minkowski inequality to easily obtain the integrability of the convolution when $G$ is a locally compact Hausdorff topological group (not necessarily $\sigma$-compact) and integration is done w.r.t. a Haar (left invariant) measure.

Suppose $G$ is not $\sigma$-compact. There is a clopen subgroup $H$ that is $\sigma$-compact, say $H=\bigcup_nK_n$ with $K_n$ compact. Since $|f|,\,|g|^p\in L_1$, the sets $A=\{f\neq0\}=\bigcup_n\{|f|>1/n\}$ and $B=\{g\neq0\}=\bigcup_n\{|g|>1/n\}$ are $\sigma$-finite w.r.t. Haar measure.

Claim I: There are $\sigma$-compact sets $A'$ and $B'$ that contain $A$ and $B$ respectively. Indeed, if $A=\bigcup_nA_n$ with $\lambda(A_n)<\infty$ then, the set $\{x\in G: A\cap (xH)\neq\infty\}$ is at most countable. Then $$A_n=\bigcup_jA\cap(x_jH)=\bigcup_{\ell,j}A\cap(x_jK_\ell)$$ This proves the claim.

Notice that if $f*g(x)=\int f(y)g(y^{-1}x)\,dy\neq0$ then $x\in AB\subset A'B'$ where for any nonempty sets $X, Y\subset G$, $X\cdot Y=\{xy:x\in X,\, y\in Y\}$. The argument above shows that $A\cdot B$ is contained in the $\sigma$-compact set $A'B'$; hence, the domain of integration $G\times G$ can be substituted by $A'\times (A'\cdot B')$, which is $\sigma$-compact and thus, $\sigma$-finite. Then, Fubini-Tonelli's theorem and all the goodness that comes with it holds; in particular, the conditions of the generalized Minkowski inequality are satisfied.

Mittens
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  • Looks good! But I don't think you need to argue using topological arguments (i.e. $\sigma$-compactness). You can just say that $A = {f\ne 0}$ and $B={g \ne 0}$ are $\sigma$-finite (by integrability, you don't need anything for this) and then apply Fubini on $A \times (AB)$. Indeed, Fubini's theorem applies to measurable functions $(x,y)\mapsto h(x,y)$ that vanish outside a rectangle with $\sigma$-finite sides. – Andromeda Apr 22 '22 at 18:58
  • The issue is to prove that $AB$ is indeed $\sigma$-finite. It is true, as I've shown it, but not so obvious. It does require the existence of the coplen subgroup $H$ – Mittens Apr 22 '22 at 18:59
  • Ah I see, good point! – Andromeda Apr 22 '22 at 19:00
  • Thanks again for the help. I think at this point I am confident enough about these technicalities to ignore them a bit. – Andromeda Apr 22 '22 at 19:03
  • This answer cleared my doubts. I was always suspicious of Harmonic analysts using Fubini's theorem in context of radon products of measures. Folland does give a little hint towards this but this answer is self contained. – NewB May 20 '23 at 09:43
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Another detail that Folland's proof does not prove is that the convolution $$(f \star g)(x) = \int_G f(y)g(y^{-1}x)dy$$ exists for a.e. $x\in G$, where $f \in L^1(G)$ and $g \in L^p(G)$ ($1 \le p < \infty$).

To see this, first note that $$\int_{G}\left(\int_G |g(y^{-1}x)|^p|f(y)|dy\right)dx = \int_G\left(\int_G |g(y^{-1}x)|^p dx\right) |f(y)|dy = \int_G \|g\|_p^p |f(y)|dy = \|g\|_p^p\|f\|_1 < \infty$$ from which it follows that $$\int_G |g(y^{-1}x)|^p |f(y)|dy <\infty$$ for almost every $x$. But then note that for almost every $x$ we have by Hölder's inequality (applied to the measure $|f(y)|dy$ and the functions $y \mapsto |g(y^{-1}x)|$ and $y \mapsto 1$) that $$\int_G |f(y)||g(y^{-1}x)|dy \le \left(\int_G |g(y^{-1}x)|^p|f(y)|dy\right)^{1/p} \left(\int_G |f(y)|dy\right)^{1/'p}< \infty$$ where $1/p+1/p' = 1$.

Andromeda
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