# Weak convergence of function to $0$

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?

• 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. – Ian May 22 at 14:32
• Should the integral be done on $[0,n]$ or on $[0,1]$ as you wrote? – Bremen000 May 22 at 14:34
• 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. – Ian May 22 at 14:55
• Related ... see "Riemann-Lebesgue lemma" en.wikipedia.org/wiki/Riemann–Lebesgue_lemma – GEdgar May 22 at 15:55
• @Ian do you think it is ok this way? – 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 $$L^\infty\subset L^2$$.

• Thank you! I edited my question using some of your answer, do you think it is ok? – 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 $$\int f_ng=\int^{y_{k_{1}}}_{a}f_ng+\int^b_{x_{k_{2}}}f_ng\le y_{k_{1}}-a+b-x_{k_{2}}\le \epsilon/2+\epsilon/2=\epsilon.$$