1
$\begingroup$

For any function $f\colon\mathbb{R}^n\to\mathbb{R}$, set: $\tau_hf(x):=f(x+h)$, $x,h\in\mathbb{R}^n$. Consider the following finite measure on $\mathbb{R}^n$: $$\mu(A):=\int_A\frac{1}{1+|y|^{n+a}}\,dy$$ where $a>0$ is given. I want to prove that:

  1. $\tau_h f\in L^1(\mathbb{R}^n,d\mu)$, $\forall f\in L^1(\mathbb{R}^n,d\mu).$
  2. For any $f\in L^1(\mathbb{R}^n,d\mu)$, we have that: $$||\tau_h f-f||_{L^1(\mathbb{R}^n,d\mu)}\to 0,\quad\text{for}\quad h\to0.$$

Since $\mu$ is a finite Borel measure then $\mu$ is regular, so $C_c(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n,d\mu)$. My claims are true for the functions of $C_c(\mathbb{R}^n)$. From here I can't go on anymore. Any help is appreciated.

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ 1. you can check that for any $h\in\mathbf{R}^n$, $y\mapsto \frac{1+|y|^{n+a}}{1+|y-h|^{n+a}}$ is bounded and conclude. 2. the second point is done as usual by the density argument (done in any textbooks) $\endgroup$
    – Ayman Moussa
    Commented Nov 12, 2020 at 13:20
  • $\begingroup$ @AymanMoussa How i can prove that, for any $h\in\mathbb{R}^n$, the function $y\mapsto\frac{1+|y|^{n+a}}{1+|y-h|^{n+a}}$ is bounded on the whole $\mathbb{R}^n$? $\endgroup$
    – inoc
    Commented Nov 12, 2020 at 14:59
  • $\begingroup$ Well it's continuous and tends to 1 when $|y|\rightarrow +\infty$. I think your question has nothing to do in MO and should have probably be posted on math.stackexchange.com $\endgroup$
    – Ayman Moussa
    Commented Nov 13, 2020 at 5:47

0

Reset to default

You must log in to answer this question.