Prove the inequality $\|\mu\ast\nu\|\leq \| \mu\|\| \nu\|$ where $\ast$ denotes convolution of measures in $M(G)$.

Question

I am working through Folland's $\textit{A Course in Abstract Harmonic Analysis}$, Second Edition, on the section which introduces convolution of measures (section 2.5, pg 54).

We define the $\textbf{convolution}$ of $\mu$ and $\nu$ by $\mu* \nu\in M(G)$ where $I(\phi)=\int\phi \,d(\mu* \nu)=\int\int \phi(xy)\,d\mu(x)d\nu(y)$.

We have that $\|\mu\ast\nu\|\leq \| \mu\|\| \nu\|$, which makes $M(G)$ under this convolution a Banach algebra.

I am struggling to show this inequality holds. I assume $\| \mu\|$ is the total variation norm, i.e. $\| \mu |\:=|\mu|(G)$. (I take this from Rudin, $\textit{Real and Complex Analysis}$, Third Ed., page 116.

I am struggling to work out what we can write $\|\mu\ast\nu\|$ as. So far I have: $\|\mu\ast\nu\|=|\mu\ast\nu|(G)=\sup\sum^{\infty}_{i=1}|(\mu\ast\nu)(E_i)|$, where ${E_i}$ is a partition of $G$ and the supremum is taken over all partitions of $G$.

There is no way to really decompose this again into a way that relates it to $\sup\sum^{\infty}_{i=1}|\mu(E_i)|.|\nu(E_i)|$.

Can anyone provide a hint as to what needs to be done to obtain the inequality from here? Thanks

Answer 1

Let us recall $M(G)$ is the space of complex Radon measures on $G$ and $G$ is a locally compact, Hausdorff group in Folland's book.

By definition, $\mu * \nu$ is given by integration according to the following rule: \begin{equation*} \int_{G} f(g) \, (\mu * \nu)(dg) = \int_{G} \int_{G} f(g h) \mu(dg) \nu(dh) \quad (f \in C_{0}(G)). \end{equation*} Therefore, \begin{equation*} \sup \left\{ \int_{G} f(g) \, (\mu*\nu)(dg) \, \mid \, f \in C_{0}(G), \, \, \|f\|_{\text{sup}} \leq 1 \right\} \leq \int_{G} \int_{G} |f(gh)| |\mu|(dg) \, |\nu|(dh) \leq |\mu|(G) |\nu|(G). \end{equation*} Note that, for each $\eta \in M(G)$, we have \begin{equation*} |\eta|(G) = \sup \left\{ \int_{G} f(g) \, \eta(dg) \, \mid \, f \in C_{0}(G), \, \, \|f\|_{\text{sup}} \leq 1 \right\}. \end{equation*} (See Theorem 6.19 in Rudin's Real and Complex.) Therefore, the above (with $\eta = \mu * \nu$) gives $|\mu * \nu|(G) \leq |\mu|(G) |\nu|(G)$.

Alternatively, in some situations (e.g. in $\mathbb{R}^{d}$), it is possible to prove that $\mu * \nu$ is given explicitly by \begin{equation*} (\mu * \nu)(E) = \int_{G} \mu(E h^{-1}) \, \nu(dh). \end{equation*} (I don't know if this is true in this generality, but that's a different question...) In that case, we find \begin{equation*} |(\mu *\nu)(G)| \leq \int_{G} |\mu(G h^{-1})| \, |\nu|(dh) \leq |\mu(G)| |\nu|(G). \end{equation*}

Answer 2

Given $\mu,\nu$, define $\lambda$ as follows: $$\lambda(E):=\mu\times \nu(\{(x,y)\in G^2:xy\in E\})$$ This is a Borel measure and it clearly satisfies $|\lambda|(X)\le |\mu|(X)|\nu|(X)$: Indeed, one has $$\lambda(E)=\iint_{xy\in E}d\mu d\nu$$ Using the polar decomposition $d\mu=h_1d|\mu|, d\nu=h_2d|\nu|$, we get that $$|\lambda(E)|\le \iint_{xy\in E}d|\mu|d|\nu|=\int_G |\mu|(Ey^{-1})d|\nu|$$ And so$$||\lambda||=\sup \sum |\lambda(E_n)|\le \int_G \sum |\mu|(E_ny^{-1})d|\nu|=\int_G ||\mu||d|\nu|=||\mu||\cdot ||\nu||$$

Now, we claim that this equation holds for all $f\in C_0(G)$: $$\int_G fd\lambda=\int_G\int_G f(xy)d\mu(x)d\nu(y)$$ To prove the result, notice that it holds for simple functions with compact support and use their density. Now, $\mu*\nu$ satisfies the same relation. Thus they define the same operator on $C_0(G)$, and by uniqueness in the Riesz-Markov-Kakutani theorem they must be equal. The result follows.