Let be $f \in L^1[0,1]$, then it applies $ \int_0^1 \exp(2i\pi xk)f(x n)\,dx=0$ for $n,k\in \mathbb{N}$ with $0<k<n$.
Ideas: f can be extended to a function on $\mathbb{R}$ with period $1$, hence $f$ can be approximated by a trigonometric polynomial with respect to $L^1$-Norm $\|\|_1$. Since it applies $\|f\|_1:=\int_0^1 |f| d\lambda$ and $|\exp(2i\pi xk)f(x)|=|f(x)|$ for any$f\in L^1[0,1]$, we have $\|f\|_1=\|\ g\cdot f\|_1$, where $g(x):=\exp(2i\pi xk)$. In my opinion it seems that $L^1$-Norm $\|\|_1$ is too "coarse" for being helpful in that proof, but I see no other way.
Thanks for the help
bests
bjn
edit: Sorry for the messed up post, I hope now is all okay.
Here's what I read between the lines of your question:
With the above additions, it is true that $$ \int_0^1 \exp(2i\pi xk) f(x n)\,dx =0 ,\quad 0<k<n $$ The reason is that each value of $f$ is repeated $n$ times (periodicity), each time being multiplied by values of the exponential function that add up to $0$. To make this formal, split the integral into $n$ parts and do some substitution: $$ \begin{split} \int_0^1 \exp(2i\pi xk) f(x n)\,dx & = \frac{1}{n} \int_0^n \exp(2i\pi t k/n) f(t)\,dt \\ & = \frac{1}{n} \sum_{j=0}^{n-1} \int_{j}^{j+1} \exp(2i\pi tk/n) f(t)\,dt \\ & =\frac{1}{n} \sum_{j=0}^{n-1} \int_{0}^{1} \exp(2i\pi (t +j)k/n) f(t)\,dt \\ & =\frac{1}{n} \sum_{j=0}^{n-1} \exp(2i \pi j k/n) \int_{0}^{1} \exp(2i\pi t k/n) f(t)\,dt \end{split} $$ The punch line is that the integral does not involve $j$ anymore, and the exponentials sum up to zero (use the geometric sum formula).
Upon request, a proof using the density of trigonometric polynomials in $L^1$. For every $\epsilon>0$ there is a polynomial $$p(x) = \sum_{m=-M}^M \exp(2\pi i m x)$$ such that $\|p-f\|_{L^1}<\epsilon$. By the integral triangle inequality, $$ \left|\int_0^1 \exp(2i\pi xk) f(x n)\,dx - \int_0^1 \exp(2i\pi xk) p(x n)\,dx\right| <\epsilon $$ Direct computation shows that $$\int_0^1 \exp(2i\pi xk) p(x n)\,dx =0 $$ (just plug in the sum and simplify: you get a bunch of exponentials, each of which integrates to zero.) Conclusion follows.