Given $u \in L^p(\Bbb R)$, $1<p<\infty$, show that the following statements are equivalent:
$u\in W^{1,p}(\Bbb R)$
$\exists c>0 $ such that for $\forall h \in R$ the following inequality takes place: $$ \|u(x+h)-u(x-h)\|_{L^p(\Bbb R)}\le c|h| $$
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Can you tell me if this is right, please?
- $\forall x,h \in\Bbb R$:
$$ u(x+h)-u(x-h)=\int_{x-h}^{x+h}u'(t)dt $$ By the change of variable: $t=sh+x-h$ ($dt = hds$ i.e. $s=\left(\frac{t-x+h}{h}\right)$, for $t=x-h: s=0$, for ,$t=x+h$, $s=2$) we have: $$ \begin{split} u(x+h)-u(x-h) & = h\int_o^2 u'(x+sh-h) ds\\ \implies |u(x+h)-u(x-h)| &\le |h|\int_0^2|u'(x+sh-h)|ds \end{split} $$ From Holder's Inequality: $$ \begin{split} |u(x+h)-u(x-h)|^p &\le|h|^p\int_0^2|u'(x+sh-h)|^pds \\ &\Downarrow\\ \int_R|u(x+h)-u(x-h)|^pdx & \le|h|^p\int_Rdx\int_0^2|u'(x+sh-h)|^pds\\ & \le|h|^p\int_0^2ds\int_R|u'(x+sh-h)|^pdx \\ & =2|h|^p\int_R|u'(x+sh-h)|^pdx\\ &\Downarrow \\ \int_R|u'(x+sh-h)|^pdx & =\int_R|u'(y)|^pdy \,\,s\in(0,1) \end{split} $$ - Getting $z\in C_c^1, \forall h \in \Bbb R:$ $$ \int_R(u(x+h)-u(x-h))z(x)dx=\int_Ru(x)(z(x-h)-z(x+h))dx $$ By applying Holder's inequality: $$ \left|\int_R(u(x+h)-u(x-h))dx\right|\le c|h|\,\|z\|_{L^{p'}(\Bbb R)} $$ and multiplying with $1/h$ and letting $h\to 0$: $$ \left| \int_R uz^{\prime}\right| \le c\|z\|_{L^{p'}(\Bbb R)} \implies u \in W^{1,p} $$