As the title says im interested in knowing if $\int_{\mathbb{T}} f(x)dx = \int_{\mathbb{T}} f(x-x_0) dx$ for $f \in C_0^{\infty}( \omega), \omega\subset \mathbb{T}$ and for any $x_0 \in \mathbb{T}$ the one-dimension torus. I know that this result is true over the real line, but i would like to know if i can have the same in the torus with the restriction of $f$ that i have.
Hope u can help me. Thank you in advance
I do not know, where exactly you struggle and thus I give the maybe overly-detailed computation below.
We have that $\mathbb{T} = [0,1]/{\sim}$, where $0\sim 1$. Hence, $$ \int_{\mathbb{T}} f(x) dx = \int_{0}^1 f(x) dx. $$
Now you could define $T: L^{1}(\mathbb{T}) \to L^{1}_{loc}(\mathbb{R})$ by $T(g)(x+k) := g(x)$ for all $k\in \mathbb{Z}$ and all $x \in [0,1].$ One computes that $T(f(\cdot - x_0)) = f(x+k-x_0)$ for a $k$ such that $x+k-x_0 \in [0,1]$ and therefore $T(f(\cdot - x_0)) = T(f)(\cdot - x_0)$.
We have $$ \int_{\mathbb{T}} g(x) dx = \int_{0}^1 T(g)(x) dx $$ and hence $$ \int_{\mathbb{T}} f(x - x_0) dx = \int_{0}^1 T(f(\cdot - x_0))(x) dx = \int_0^1 T(f)(x - x_0) dx = \int_{-x_0}^{1-x_0} T(f)(x) dx. $$ Clearly, by the periodicity of $T(f)$ we have that $$ \int_{-x_0}^{0} T(f)(x) dx = \int_{-x_0}^{0} T(f)(x+1) dx = \int_{1-x_0}^{1} T(f)(x). $$ In other words $$ \int_{-x_0}^{1-x_0} T(f)(x) dx = \int_{0}^{1} T(f)(x) dx = \int_{0}^{1} f(x) = \int_{\mathbb{T}} f(x). $$ The proof works for $f \in L^1(\mathbb{T})$ so also for $f \in C^\infty_0(\mathbb{T})$ as requested.