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.

Thakning in advance.


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 $1<p<\infty$, 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 $$\lim_{n\to\infty}(f_n,g)= \int_0^Tf(nx)g(x)dx=\lim_{n\to\infty} \frac{1}{T}\int_0^Tf(x)dx\cdot\int_0^Tg(x)dx =(\bar{f},g)$$

  • $\begingroup$ Thank you very much for your reply! I just saw your answer. $\endgroup$ – passenger Jan 22 '18 at 21:20
  • $\begingroup$ @passenger your question have been here since couple of year if ever I saw it before I $\endgroup$ – Guy Fsone Jan 22 '18 at 21:22
  • $\begingroup$ I know. I forgot it completely, that's why I was surprised to see an answer. Anyway's, thank's once more for your time! $\endgroup$ – passenger Jan 22 '18 at 21:26
  • $\begingroup$ 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? $\endgroup$ – Zack Ni Dec 18 '18 at 2:22
  • $\begingroup$ @ZackNi check the link before. the answer is there. math.stackexchange.com/questions/105258/… $\endgroup$ – Guy Fsone Dec 19 '18 at 6:57

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