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?


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 $M \to \infty$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.