Hi! following the thread of Bounded linear operators that commute with translation
I can't prove that $\left\|m\right\|=\left\|T\right\|$
I have only one inequality:
$\left\|T(f)\right\|_{L^1}\leq \left\|f\right\|_{L^1}\left\| \mu\right\|$. Then, $\left\|T\right\|\leq \left\|\mu\right\|$
For the second, I thought of taking $f =\delta$ Dirac delta function which satisfies $\int_{\mathbb{R}^n}\delta(x)=1$ and $\delta\ast \mu=\mu$ to have to $\left\|T\right\|\geq \left\|T(\delta)\right\|_{L^1}=\left\|\delta\ast \mu\right\|_{L^1}=\left\|\mu\right\|_{L^1}$
then $\left\|T\right\|$ is a upper bound for $\left\|\mu\right\|_{L^1}$ then $\left\|\mu\right\|\leq \left\|T\right\|$
But I don't know if the above is legal.
Actualization 1:
Is the following valid? In Bounded linear operators that commute with translation have this:
$\lim_{k} \int f(y)T\phi_{\epsilon_k}(x-y)dy=\int f(y)\mu(x-y)dy$ with $x=2y$ $\lim_{k} \int f(y)T\phi_{\epsilon_k}(y)dy=\int f(y)\mu(y)dy$ then
\begin{align*} &\left\|\mu\right\|=\sup_{|f|_{L^1}\leq 1} \left|\int f(x)d\mu(x)\right|\\ &=\sup_{|f|_{L^1}\leq 1}\lim_{k} \left|\int f(x) T\phi_{\epsilon_k}(x)dx\right|\\ &\leq \int|T\phi_{\epsilon_k}(x)|dx\\ &=|T\phi_{\epsilon_k}|_{L^1}\leq \left\|T\right\| |\phi_{\epsilon_k}|_{L^1}\\ &=\left\|T\right\| \end{align*} Therefore $\left\|\mu\right\|\leq \left\|T\right\|$.
It is correct?
This argument is based on the demonstration of the Grafakos Classical Fourier Analysis book. Theorem 2.5.8. (characterization of multipliers for p = 1), equation 2.5.14 page 153
pd: $d\mu(x)=\mu(x)dx$?
Actualization 2.
I am trying to understand the demonstration of link Bounded linear operators that commute with translation. I have the following with some doubts:
$L^1:=L^1(\mathbb{R}^n)$
Let $\phi\in L^1$ such that $\int_{R^n}\phi(x)dx=1$ and, for any $\epsilon>0$,\ $\phi_{\epsilon}:=\epsilon^{-n}\phi(x/\epsilon)$.
Is easy to see that $\|\phi_{\epsilon}\|_{1}=\|\phi\|_{1}$
lemma. Let $\phi\in L^1$, $\|\phi\|_{1}=1$. For any $f\in L^p,\ 1\leq p<\infty,\ \lim_{\epsilon\to 0}\|f\ast \phi_{\epsilon}-f\|_{p}=0$.
Observation: $\{\phi_{\epsilon}\}_{\epsilon>0}$ is an aproximation identity.
By the lemma, $\lim_{\epsilon\to 0} \int_{R^n}f\ast \phi_{\epsilon}(x)dx=\lim_{R^n}f(x)dx$.
Now, as $\|\phi_{\epsilon}\|_{1}=1$ then $\|T\phi_{\epsilon}\|_{1}$ is bounded in $L^1$ because $T$ is bounded.
Because $L^1$ is naturally embedded in the space of finite Borel measures, wich is the dual of the space $C_{0}:=C_{0}(R^n)$ of continuous functions that tend to zero at infinity, we obtain that the family $T\phi_{\epsilon}$ lies in a fixed multiple of the unit ball of $(C_{0})^{\ast}$. By the Banach-Alaoglu theorem, this is a $weak^{\ast}$ compact set. Therefore, some subsequence of $T\phi_{\epsilon}$ converges in the $weak^{\ast}$ topology to a measure $\mu$. That is, for some $\epsilon_{k}\to 0$, and all $g\in C_{0}$, we have
$$\lim_{k\to\infty}\int_{R^n}g(x)T\phi_{\epsilon_{k}}(x)dx=\int_{R^n}g(x)\mu(x)dx$$ (Here I used $d\mu(x)=\mu(x)dx$ that which I don't know if it's true ...
Now, as $T$ is linear, commute with traslations, in addition of the fact that $C_{0}$ is dense in $L^1$ and $T\phi_{\epsilon_{k}}\to \mu$, $weak^{\ast}$ convergence, it has:
\begin{align*} f\ast \mu(x)&=\int_{\mathbb{R}^n}f(y)\mu(x-y)dy\\ &=\lim_{k\to\infty} \int_{\mathbb{R}^n}f(y)(T\phi_{\epsilon_{k}})(x-y)dy\\ &=\lim_{k\to\infty} \int_{\mathbb{R}^n}f(y)\tau_{y} (T\phi_{\epsilon_{k}}(x))dy\\ &=\lim_{k\to\infty} \int_{\mathbb{R}^n}f(y) T(\tau_{y}\phi_{\epsilon_{k}})(x)dy\\ &=\lim_{k\to\infty}\int_{\mathbb{R}^n}f(y)T(\phi_{\epsilon_{k}}(x-y))dy\\ &=\lim_{k\to\infty}T\left(\int_{\mathbb{R}^n}f(y)(\phi_{\epsilon_{k}}(x-y))dy\right)\\ &=Tf(x) \end{align*}
In the above, I do not know if the following is correct:
$$\text{If } \lim_{k\to\infty}\int_{R^n}f(y)T\phi_{\epsilon_{k}}(y)dy=\int_{R^n}f(y)\mu(y)dy$$ for any $f\in L^1$ (because $C_0$ is dense in $L^1$) Implies $$\lim_{k\to\infty}\int_{R^n}f(y)T\phi_{\epsilon_{k}}(x-y)dy=\int_{R^n}f(y)\mu(x-y)dy ?$$
thanks
