Let $\mathbb{T}^n := (\mathbb{R}/\mathbb{Z})^n$ be the $n$-dimensional torus and consider $L^1(\mathbb{T}^n)$.
Then, any element $f \in L^1(\mathbb{T}^n)$ can in fact be regarded as a locally integrable function on whole $\mathbb{R}^n$.
Now, let $Q$ be any bounded hypercube in $\mathbb{R}^n$. Then, is there some universal $C_n$ depending only on $n$ such that \begin{equation} \frac{1}{\lvert Q \rvert } \int_{Q} \lvert f(x) \rvert dx \leq C_n \lVert f \rVert_{L^1(\mathbb{T}^n)} \end{equation} holds for "all" $f \in L^1(\mathbb{T}^n)$?
I strongly suspect so, but cannot really prove anything from scratch. In fact, I think the same bound would be still valid even if we generalize $Q$ to all "compact measurable sets" in $\mathbb{R}^n$.
Could anyone please help me?
Your bound would need to depend on the size of $Q$, and the constant may become very large if $|Q|\to 0$. Indeed, if $f$ has a singularity, then your desired inequality asserts that all averages of $f$ near the singularity are bounded by the $L^1$ norm of $f$, which clearly can’t occur since the average value of a function near a singularity is generally large. For an explicit example take $f(x) = 1/(2\sqrt{x})$ on $Q = [0,\epsilon]$ then $$\frac1{|Q|}\int_Q |f| = \frac{\sqrt{\epsilon}}{\epsilon} = |Q|^{-1/2}$$ which blows up as $\epsilon \to 0$.
Morally speaking, the left hand side should approximate the local behavior of $f$ as $|Q|\to 0$ (by the Lebesgue differentiation theorem), so you should not expect to be able to bound local behavior by global information like the $L^1$ norm.