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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 $1<p<\infty$. 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?

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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 $M \to \infty$.

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  • $\begingroup$ Thanks! I missed this equivalent definition of uniform integrability. It is indeed invariant against the transformation here. $\endgroup$ – daw Feb 7 '20 at 9:19

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