Let $u\in W^{1,p}(R)$ be given, $1\leq p<\infty$. We define $$ \tau_h(u)(x):=\frac{u(x+h)-u(x)}{h} $$ be the difference quotient.
We all know that up to a subsequence $\tau_h(u)\to u'$ in the weak sense and $\|\tau_h(u)\|_{L^p}\to \|u'\|_{L^p}$.
But my professor claims that $$ \|\tau_h(u)-u'\|_{L^p(R)}\to 0 $$ as $h\to 0$.
In the beginning I was trying to use approximation to do it. Let $u_\epsilon$ be sequence of $C_c^\infty$ function such that $u_\epsilon\to u$ in $W^{1,p}$. Of course, we have $$ \|\tau_h(u_n)-u_\epsilon'\|_{L^p(R)}\to 0 $$ Hence I was trying to use $$ \|\tau_h(u)-u'\|_{L^p(R)}\leq \|\tau_h(u)-\tau_h(u_n)\|_{L^p(R)}+\|\tau_h(u_n)-u_n'\|_{L^p(R)}+\|u_n'-u'\|_{L^p(R)}$$
However, I can not go future from here because I can NOT control the rate of convergence of term $$ \|\tau_h(u)-\tau_h(u_n)\|_{L^p(R)}$$ That is, for any $\epsilon>0$ be given, I can only choose $n$ large enough such that $$ \|\tau_h(u)-\tau_h(u_n)\|_{L^p(R)}<\epsilon\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ for EACH fixed $h$. Next I have to choose $h$ small enough such that $$\|\tau_h(u_n)-u_n'\|_{L^p(R)}<\epsilon$$ However, now I lose the central of the part $$ \|\tau_h(u)-\tau_h(u_n)\|_{L^p(R)}$$ It may happen that as $h$ gets small, I have to choose a bigger $n$ to maintain $(1)$ hold...and we fall into kind of circular argument...
Any help is really welcome!
Update: for case $p=2$ here is a quick prove. Since we know that, up to a subsequence, $\tau_h(u)\to u'$ in the weak sense and $\|\tau_h(u)\|_{L^2}\to \|u'\|_{L^2}$. Hence, by $L^2$ is a Hilbert space we actually have $$ \|\tau_h(u)-u'\|_{L^2(R)}\to 0. $$
But the same argument can not hold if $p\neq 2$, since it is not a Hilbert space.
You can use the fact that $L^p$ spaces are a Radon-Riesz spaces, for p>1. and the case $p=1$ i don't have a proof but you can use the fact that each of the $L^p$-spaces, $1 ≤ p < ∞$, has the property that each sequence on the unit sphere that converges almost everywhere converges also in norm.