Hello again. I have some doubts in this theorem. I am trying to see all the details to understand it. I have the following
equation 2.5.13 is:
$<g,f_{\epsilon_{k}}*u>\to <g,\mu>$.
Affirmation. $<g,f_{\epsilon_{k}}*u>=<u,\tilde{f}_{\epsilon_{k}}*g>$
Proof:$<g,f_{\epsilon_{k}}*u>: =\int g(x)(f_{\epsilon_{k}}*u)(x)dx\\ =\int g(x)(\int f_{\epsilon_{k}}(y)u(x-y)dy)dx\\ =\int g(x) (\int\tilde{f}_{\epsilon_{k}}(-y)u(x-y)dy)dx\\ =\int g(x) (\int\tilde{f}_{\epsilon_{k}}(y)u(x+y)dy)dx\\ =\int g(z-y) (\int\tilde{f}_{\epsilon_{k}}(y)u(z)dy)dz\\ =\int u(z) (\int\tilde{f}_{\epsilon_{k}}(y)g(z-y)dy)dz\\ =\int u(z) (\tilde{f}_{\epsilon_{k}}*g)(z)dz\\ =<u,\tilde{f}_{\epsilon_{k}}*g>$.
Affirmation 2. $<u,\tilde{f}_{\epsilon_{k}}*g>=<u,f_{\epsilon_{k}}*g>$
proof: Similary to affirmation 1.
Exists a subsequence $f_{\epsilon_{k}}*u\subset f_{\epsilon}*u$ such that $f_{\epsilon_{k}}*u\to \mu$, for some $\mu\in \mathcal{M}$ (Borel finite measure).
Here my main question. $f_{\epsilon_{k}}*u,\ \mu$ are in $(C_{0}(R^n))^{\ast}$ right?, i.e. they are continuous linear functionals i.e. $f_{\epsilon_{k}}(g)\to \mu(g)$ for any $g\in C_{0}(R^n)$. Right?
$f_{\epsilon_{k}}*u\to \mu$ weak star convergence.
This is, $\| f_{\epsilon_{k}}*u-\mu\|_{\mathcal{M}}\to 0$. Right?
$\| f_{\epsilon_{k}}*u-\mu\|_{\mathcal{M}}:=\sup_{g\in C_{0}(R^n)}\left|\int g(x)[(f_{\epsilon_{k}}*u)(x)-\mu(x)]dx\right|$ ??
For this context, what is the norm $\|\mu\|_{\mathcal{M}}$?
Why in Grafakos, $f_{\epsilon_{k}}*u\to \mu$ weak star convergence implies $\lim_{k} \int g(x)(f_{\epsilon_{k}}*u)(x)dx=\int g(x)d\mu(x)$?
pd: $d\mu(x):=\mu(x)dx$?
actualization 1:
Theorem (Riesz Representation) If $X$ locally compact space and $\mu\in \mathcal{M}(X)$, define $F_{\mu}:C_{0}(X)\to \mathcal{F}$ by $F_{\mu}=\int f d\mu$. Then $F_{u}\in (C_{0})(X))^{\ast}$ and the map $\mu\mapsto F_{\mu}$ is an isometric isomorphism of $\mathcal{M}(X)$ onto $(C_{0}(X))^{\ast}$.
Affirmation: $L^1(R^n)$ can be identified as a subspace of $\mathcal{M}(R^n)$ (finite borel measure) by the isometry $f(x)\to f(x)dx$, $dx$ leebesgue measure. This affirmation, appears in Stein, Singular Integrals and differenciability properties.
Now, as $f_{\epsilon_{k}}*u\in L^1$ then $f_{\epsilon_{k}}*u dx \in \mathcal{M}(R^n)=(C_{0}(R^n))^{\ast}$ and $\mu \in\mathcal{M}(R^n)=(C_{0}(R^n))^{\ast}$. therefore by Riesz Representation, $f_{\epsilon_{k}}*u,\ \mu$ are identified by $\int g(x)(f_{\epsilon_{k}}*u)(x)d\mu$ and $\int g(x)d\mu$ for any $g\in C_{0}(R^n)$. It is correct?
As $\lim_{k} f_{\epsilon_{k}}*u=\mu$, *-weak convergence, for the above, $\lim_{k} \int g(x)(f_{\epsilon_{k}}*u)(x)dx=\int g(x)d\mu(x)$.
It is correct?