Weak convergence of oscillating functions in $L^1(0,1)$

Let $$I:=(0,1)$$. Let $$f\in L^1_{loc}(\mathbb R)$$ be a periodic function with period $$1$$. Assume $$f|_I \in L^1(I)$$ and $$\int_I f(x) dx=0$$. Define the oscillating sequence $$f_n(x):=f(nx).$$ This sequence is bounded in $$L^1(I)$$ since $$\|f_n\|_{L^1(I)} = \|f\|_{L^1(I)}$$ due to periodicity.

Then it is well-known that $$(f_n)$$ converges weakly to zero in $$L^p(I)$$ for all $$1. It converges weakly-star to zero in $$L^\infty(I) = L^1(I)^*$$ and in $$L^1 \subset M(I) = C(\bar I)^*$$.

The proofs of these statements use density of characteristic functions of intervalls (1) or of continuous functions (2) in the respective dual spaces. If $$f\in L^p_{loc}(\mathbb R)$$ for some $$p>1$$ then weak convergence in $$L^1(I)$$ follows from embedding.

However, these techniques do not work to show weak convergence in $$L^1$$, as its dual space is $$L^\infty$$, and these density arguments cease to work.

Another possibility is to use Dunford-Pettis theorem by showing that the sequence $$(f_n)$$ is uniformly integrable. But I do not see how to achieve this.

My question is: Is it possible to prove $$f_n \rightharpoonup 0$$ in $$L^1(I)$$ or is there a counterexample of a periodic function $$f$$, where this fails?

As you noticed, it suffices to show that the sequence is uniformly integrable. There are several equivalent formulations of this. Since (as you noted) the sequence $$(f_n)_{n \in \Bbb{N}}$$ is bounded in $$L^1$$, it suffices to prove that $$\sup_n \int_0^1 |f_n(x)| \cdot 1_{|f_n(x)| \geq M} \, d x \to 0$$ as $$M \to \infty$$.
That this is indeed satisfied can be verified as follows: \begin{align*} & \int_0^1 |f_n (x)| \cdot 1_{|f_n(x)| \geq M} \, d x \\ & = \frac{1}{n} \int_0^1 n \cdot |f(n x)| \cdot 1_{|f(nx)| \geq M} \, d x \\ & = \frac{1}{n} \int_0^n |f(y)| \cdot 1_{|f(y)| \geq M} \, d y \\ & = \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(y+i)| \cdot 1_{|f(y+i)| \geq M} \, d y \\ & \overset{(\ast)}{=} \frac{1}{n} \sum_{i=0}^{n-1} \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z \\ & = \int_0^1 |f(z)| \cdot 1_{|f(z)| \geq M} \, d z. \end{align*} Here, we used the periodicity of $$f$$ at the step marked with $$(\ast)$$.
Note that the right-hand side of the above estimate is independent of $$n$$, and converges to zero as $$M \to \infty$$.