# weakly convergence of periodic functions

Let $\displaystyle I \subset \mathbb R$ be a bounded interval and $\displaystyle 1 < p \leq \infty$. Let $f \displaystyle \in L^\infty (\mathbb R)$ periodic function with $\displaystyle f(x+T)=f(x), \quad \forall x \in I$. Consider now the sequence of functions $\displaystyle f_n (x):= f(nx), \quad n \in \mathbb N$. Prove that:

(i) $\displaystyle f_n \to \frac{1}{T} \int_0^T f(x) dx$ weakly in $\displaystyle L^p (I)$ forall $\displaystyle 1<p < \infty$.

(ii) $\displaystyle f_n \to \frac{1}{T} \int_0^T f(x) dx$ weakly* in $\displaystyle L^\infty (I)$.

I thought the following about the first one: First of all we can assume that $\displaystyle \frac{1}{T} \int_0^T f(x) dx =0$, otherwise work with $\displaystyle f(x) - \frac{1}{T} \int_0^T f(x) dx =0$. Consider now an arbitrary compact sub-interval of $I$, i.e., $\displaystyle [a,b] \subset I$ . Then we have that:

$\displaystyle \int_a^b f_n(x) dx= \int_a^b f(nx) dx = \frac{1}{n} \int_{na}^{nb} f(x) dx = \frac{1}{n} \left( F(nb) -F(na) \right) \to 0$, as $n \to \infty$

where $\displaystyle F(x) = \int_0^x f(t) dt , \quad x \in I$.

Since the sub-interval $[a,b]$ was rbitrary, we can conclude that for every step function $\phi$ we have that: $\displaystyle \int_I f_n(x) \phi (x) dx \to 0$ , as $n \to \infty$. Because the step functions are dense in $L^q(I)$ (is the dual space of $L^p(I)$) we have that $\displaystyle \int_I f_n(x) g(x) dx \to 0$ as $n \to \infty$, for every $g \in L^q(I)$.

From here how can I get the desired result?

Also I would like if it is possible some hints for the (ii)

Any help would be really appreciated.

Well it merely derives from this integral involving a periodic function we have that

for all continuous function $$g\in C(I)$$ we have $$\lim_{n\to\infty}(f_n,g)= \int_0^Tf(nx)g(x)dx= \frac{1}{T}\int_0^Tf(x)dx\cdot\int_0^Tg(x)dx =(\bar{f},g)$$ Where $$(\cdot,\cdot)$$ is the dual pairing between $$L^p(I)$$ and $$L^{p'}(I)= (L^{p}(I))'$$ for $$1, for $$L^1(I)$$ and $$(L^1(I))'= L^\infty(I)$$ and similarly for $$L^\infty(I)$$ and $$L^1(I)\subset (L^\infty(I))'$$ see here and we denote the mean of $$f$$ by, $$\bar{f} =\frac{1}{T}\int_0^Tf(x)dx$$

For general function $$g\in L^{p'}(I)$$ the result is merely just a consequence of density of smooth functions in $$L^p$$ as displayed below.

Now by density argument we know if $$g\in L^{p'}(I)$$ then for fixed $$\varepsilon>0$$ there is $$g_\varepsilon \in C(I)$$ such that $$\|g-g_\varepsilon\|_{p'}<\varepsilon$$

since $$f$$ is continuous and periodic, it is bounded and therefoer By Holder inequality we have, \begin{align}|(f_n,g)-(\bar{f},g)| &=|(f_n-\bar{f},g)|\\&\le|(f_n-\bar{f},g_\varepsilon)|+|(f_n-\bar{f},g-g_\varepsilon)|\\&\le |(f_n-\bar{f},g_\varepsilon)| +\|f_n-\bar{f}\|_{p }\|g-g_\varepsilon\|_{p'} \\&\le |(f_n-\bar{f},g_\varepsilon)| +(1+T^{1/p})\|f\|_{\infty }\|g-g_\varepsilon\|_{p'} \\&\le |(f_n-\bar{f},g_\varepsilon)| +(1+T^{1/p})\|f\|_{\infty }\varepsilon \end{align}

however, we have already prove that $$\lim_{n\to\infty}(f_n,g) =(\bar{f},g_\varepsilon)\Longleftrightarrow \lim_{n\to\infty}(f_n-\bar{f}, g_\varepsilon)=0$$ we get

$$\limsup_{n\to\infty}|(f_n,g)-(\bar{f},g)| \le(1+T^{1/p})\|f\|_{\infty }\varepsilon$$ then letting $$\varepsilon \to0$$ one obtains

$$\lim_{n\to\infty}(f_n,g) =(\bar{f},g)$$

that is for all $$g\in L^{p'}(I)$$ we have $$\lim_{n\to\infty}(f_n,g)= \int_0^Tf(nx)g(x)dx=\lim_{n\to\infty} \frac{1}{T}\int_0^Tf(x)dx\cdot\int_0^Tg(x)dx =(\bar{f},g)$$

• Why in $\lim_{n\to\infty}(f_n,g)= \int_0^Tf(nx)g(x)dx=\lim_{n\to\infty} \frac{1}{T}\int_0^Tf(x)dx\cdot\int_0^Tg(x)dx =(\bar{f},g)$, the second equality follows? – Zack Ni Dec 18 '18 at 2:22