Suppose $\mu = \mathcal{L} |_{[0,1]}$ where $\mathcal{L}$ is the Lebesgue measure on the real line.

Define $$f(x):= \begin{cases} 1 & \text{if } 0 \le x < \frac{1}{2} \\ -1 & \text{if } \frac{1}{2} \le x < 1 \end{cases}$$ Periodically extended with period $1$ on the whole real line. Define, for each $n \in \mathbb{N}$, $f_n(x) = f(nx)$. How can I prove $$ f_n \rightharpoonup 0 $$ in $L^p(\mathbb{R}, \mu)$, per every $p \in [1, + \infty)$?

I should prove that, taken $p \in [1, + \infty)$ and $q$ s.t. $\frac{1}{p}+\frac{1}{q} =1$ then, for each $g \in L^{q}(\mathbb{R}, \mu)$, we have $$ \lim_{n \to + \infty} \int_0^1 f(nx)g(x) dx =0 $$

Using the change of variables formula I only obtain $$\int_0^1 f(nx)g(x) dx = \frac{1}{n} \int_0^n f(x)g(x/n)dx $$ and I am not able to conclude from this.

EDIT: Maybe I did it (using your advices) Take $\varphi \in C^{\infty}_c((0,1))$, then \begin{align*} \Biggl | \int_{\mathbb{R}} f_h(x) \varphi(x) \text{d} \mu(x) \Biggr |&= \Biggl | \frac{1}{n} \int_0^n f(x) \varphi(x/n) \text{d} x \Biggr | = \Biggl |\frac{1}{n} \sum_{i=0}^{n-1} \biggl ( \int_{i}^{i+\frac{1}{2}} \varphi(x/n) \text{d} x - \int_{i+\frac{1}{2}}^{i+1} \varphi(x/n) dx \biggr ) \Biggr |\\ & = \Biggl | \frac{1}{2n} \sum_{i=0}^{n-1} ( \varphi(x_n^{i,1}/n)-\varphi(x_n^{i,2}/n) \Biggr |= \Biggl | \frac{1}{2n} \sum_{i=0}^{n-1} \varphi'(\xi_n^i)\frac{ x_n^{i,1}-x_n^{i,2}}{n} \Biggr | \\ & \le \frac{1}{2n^2} \sum_{i=0}^{n-1} \| \varphi'\|_{L^{\infty}(0,1)} \le \frac{ \| \varphi'\|_{L^{\infty}(0,1)} }{2n} \overset{ n \to + \infty}{\longrightarrow} 0 \end{align*} By density we have also that the result holds for $\varphi \in L^q(\mathbb{R} , \mu) $ with $\frac{1}{p} + \frac{1}{q} = 1$.

Is it ok?

  • $\begingroup$ The idea is that in the integral $\int_0^1 f(x) g(x/n) dx$, $g(x/n)$ has more or less the same values on $[1/2,1]$ as it does on $[0,1/2]$ and so there is a cancellation effect. This is the same idea as the Riemann-Lebesgue lemma essentially. $\endgroup$ – Ian May 22 at 14:32
  • $\begingroup$ Should the integral be done on $[0,n]$ or on $[0,1]$ as you wrote? $\endgroup$ – Bremen000 May 22 at 14:34
  • $\begingroup$ You run the argument on $[0,1]$ (in material physics this is called a "cell problem") and then you wind up averaging the error estimates you get when you return to the full problem. $\endgroup$ – Ian May 22 at 14:55
  • $\begingroup$ Related ... see "Riemann-Lebesgue lemma" en.wikipedia.org/wiki/Riemann–Lebesgue_lemma $\endgroup$ – GEdgar May 22 at 15:55
  • $\begingroup$ @Ian do you think it is ok this way? $\endgroup$ – Bremen000 May 22 at 16:13

Notation: All functions below have period $1$.

Hints: At least for $p>1$ you can assume that $g$ is continuous, because...

Now $$\int_0^1 f(x/n)g(x-1/(2n))=\int_0^1 f((x+1/(2n))n)g(x) =-\int_0^1f(xn)g(x),$$ hence $$\int_0^1 f(xn)g(x)=\int_0^1f(xn)\frac{g(x)-g(x-1/(2n))}2\dots$$

Ah. That doesn't quite work for $p=1$ because $C[0,1]$ is not dense in $L^\infty$. But here $f_n\to0$ weakly in $L^2$ implies $f_n\to0$ weakly in $L^1$, because $L^\infty\subset L^2$.

  • $\begingroup$ Thank you! I edited my question using some of your answer, do you think it is ok? $\endgroup$ – Bremen000 May 22 at 15:45

Here is a sketch of an approach, for $p \neq \infty$ that uses the "cancellation" effect of the $f_n$. By a density argument, it is enough to prove this for the step function $g(x)=\chi_{(a,b)}(x)$ where $(a,b)\subseteq [0,1]$.

Let $0<\epsilon<1.$ For each integer $n,$ the values of $f_n$ alternate between $1$ and $-1$ on subintervals $I_k=[x_k,y_k]$ of length $1/2n$, starting with $f(x)=1$ when $x\in I_1=[0,1/2n].$ Choose $N$ large enough so that $1/2N<\epsilon/2$ and suppose $n\ge N.$

If $a$ and $b$ lie in one of these subintervals, then $|\int f_ng|=1/2n<\epsilon/2.$ If not, then without loss of generality, assume $a$ lies in one of the subintervals for which $f_n(x)=1$ and that $b$ lies in another subinterval. Then, because of cancellation, there are integers $k_1$ and $k_2$ such that $\int f_ng=\int^{y_{k_{1}}}_{a}f_ng+\int^b_{x_{k_{2}}}f_ng\le y_{k_{1}}-a+b-x_{k_{2}}\le \epsilon/2+\epsilon/2=\epsilon.$


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