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I am currently reading a book called classical and multilinear harmonic analysis. In section 1.2.2, I found that I couldn't understand the third statement of proposition 1.5 says that for any $\mu \in M(\mathbb T)$ we have $$ \phi_N \ast \mu \rightharpoonup \mu, \ \ \ N\rightarrow \infty $$ in weak-$\ast$ sense, where $\{ \phi_N\}_{N=1} ^\infty$ is any approximate identity. The book says this comes from the the first statement of this proposition and duality. However, it doesn't look that simple for me. This statement is saying that $\phi_N \ast \mu $ converges to $\mu$ for any $f \in C(\mathbb T)$, but I don't know how to interpret $\phi_N \ast \mu $ (the convolution involving a measure).Is there anyone knows how to understand it.

Murray.A
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  • So here's my thinking: they mean $\phi_N\ast\mu(x) = \int_{\Bbb T} \phi_N(xy^{-1}),d\mu(y)$. This defines a function and it is in fact continuous, which means that you can associate it to a measure (which is why you can envision it converging to $\mu$ in weak*). This is probably how you should think of it. It is very odd to me that they do not explicitly say what the convolution is. – Cameron Williams Mar 16 '16 at 16:40
  • Thanks for your help, if $\phi_N \ast \mu $ is an function on $\mathbb T$ then how to understand $\phi_N \ast \mu \rightharpoonup \mu, \ \ \ N\rightarrow \infty$, I think the weak convergence here is about point wise convergence of $f \in C(\mathbb T)$ not for $x\in \mathbb T$ – Murray.A Mar 16 '16 at 16:51
  • $\int_{\mathbb{T}} (\phi_N*\mu) f,d\theta \to \int f,d\mu$ if $N\to\infty$ for every $f\in C(\mathbb{T})$. – user90189 Mar 19 '16 at 04:08

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