Vrftsjtryk

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 $gin C(I)$ we have $$lim_{ntoinfty}(f_n,g)= int_0^Tf(nx)g(x)dx= frac{1}{T}int_0^Tf(x)dxcdotint_0^Tg(x)dx =(bar{f},g)$$
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 $gin L^{p'}(I)$ 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 $gin L^{p'}(I)$ 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
$$lim_{ntoinfty}(f_n,g) =(bar{f},g_varepsilon)Longleftrightarrow lim_{ntoinfty}(f_n-bar{f}, g_varepsilon)=0 $$
we get

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

$$lim_{ntoinfty}(f_n,g) =(bar{f},g) $$

that is for all $gin L^{p'}(I)$ we have $$lim_{ntoinfty}(f_n,g)= int_0^Tf(nx)g(x)dx=lim_{ntoinfty} frac{1}{T}int_0^Tf(x)dxcdotint_0^Tg(x)dx =(bar{f},g)$$