Convolution inequality $\|f\star g\|_p \le \|f\|_1 \|g\|_p$ on locally compact group

Question

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 $G$.

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?

Answer 1

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 $$|f*g(x)|\leq\|g\|_\infty\int_G|f(xy)|\,\lambda(dy)=\|g\|_\infty\|f\|_1$$

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

Answer 2

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 $1/p+1/p' = 1$.

Answer 3

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.