We define weighted Lebesgue norm as follows:
$\|f\|_{L_{w}^p}^p= \int_{\mathbb R^d} |f(x)|^p w(x) dx$
where $w$ is some nonnegative weight function (e.g., $ w(x)= (1+|x|^2)^{1/2}$ )
Fix $y\in \mathbb R^d,$ $T_yf(x)=f(x-y)$ (translation operator), we know that $\|T_yf\|_{L^p}=\|f\|_{L^p}$
My Question is: Can we say that $\|T_yf\|_{L_w^p}=\|f\|_{L_w^p}$?
I don't believe so, here is my proposition.
We know
$$\|T_yf\|_{L^p}=\|f\|_{L^p}$$
This is because the Lebesgue Measure is translationally invariant.
i.e. $$E\in \mathcal{M}\implies \mu(x+E)=\mu(E)$$
Where $\mathcal{M}$ is the sigma algebra of measurable sets.
This does not hold for all measures.
So in our case let's consider the following counterexample.
Let $E\in \mathcal{M}$ and say $\mu(E)<\infty$ and let's also assume that E is a bounded set.
Consider $\chi_E$ the characteristic function defined on E. I think we can agree that $$\chi_E\in L^p$$ and $$\chi_E\in L^p_w$$ If we apply $T_y$ to our function then $\|\cdot\|_{L^p_w}$ will not be preserved because our weight varies with position and so the measure of E will not be translation invariant.
This is just what my intuition says, perhaps someone else can elaborate a little more.
Take care.