Showing $f$ has a weak derivative in $L^p$

Question

Let $1<p<\infty$ and $f\in L^p(\mathbb{R})$. Furthermore, assume that $$\sup_{0<|h|<1}\int \left|\frac{f(x+h)-f(x)}{h}\right|^p\,dx<\infty.$$ I would like to show that $f$ has a weak derivative $g$ that belongs to $L^p$, the former meaning that $$\int g\psi \,dx=-\int f\psi'\,dx$$ for all test functions $\psi\in C_c^\infty(\mathbb{R})$. By Fatou, we have $$\int \liminf_{h\to 0} \left|\frac{f(x+h)-f(x)}{h}\right|^p\,dx\le\liminf_{h\to 0} \int \left|\frac{f(x+h)-f(x)}{h}\right|^p\,dx<\infty.$$

At this point, I would like to conclude that $$\int|f'|^p<\infty$$ but I know that this doesn't necessarily follow from the inequality that Fatou gives.

Answer 1

Let $$\sup_{0<|h|<1}\int \left|\frac{f(x+h)-f(x)}{h}\right|^p\,dx=:M.$$ Consider a mollification on $f_\varepsilon$ of $f$. Since the $L^p$ norm of a mollification of a function is less than or equal to the $L^p$ norm of the function, you have that $$\sup_{0<|h|<1}\int \left|\frac{f_\varepsilon (x+h)-f_\varepsilon (x)}{h}\right|^p\,dx\le M$$ Now if you use your trick with Fatou’s, which now works because $f_\varepsilon$ is smooth, you get that $$\int |f’_\varepsilon|^pdx\le M.$$ It follows that the sequence $\{f_\varepsilon\}$ is bounded in $W^{1,p}$. Since $p>1$, this space is reflexive, and so you can find a sequence that converges weakly to a function $g$ in $W^{1,p}$. But since the mollifications converge to $f$ in $L^p$, you get that $f=g$.

Answer 2

Let $$<u_h,\phi> = \int \frac{f(x+h)-f(x)}{h} \phi$$ where $\phi$ is a test function.

$u_h$ is a family of bounded linear functional. By corollary 19.17 https://mathweb.ucsd.edu/~bdriver/231-02-03/Lecture_Notes/weak-derivatives.pdf i.e., by banach-Alaoglu theorem, for some $L_p$ function $u$, for some sequence $h_k \rightarrow 0$, $$ \lim_{h_k \rightarrow 0} <u_{h_k},\phi> = <u,\phi>$$

Now we have $$\int \frac{f(x+h)-f(x)}{h} \phi(x) =\frac{1}{h} \int f(x+h) \phi(x)- \frac{1}{h} \int f(x) \phi(x)=\frac{1}{h} \int f(y) \phi(y-h)- \frac{1}{h} \int f(y) \phi(y) = \int \frac{\phi(x-h)-\phi(x)}{h} f(x)$$

Hence $$<u,\phi> = \lim_{h_k \rightarrow 0} <u_{h_k},\phi> = \lim_{h_k \rightarrow 0} \int \frac{f(x+h_k)-f(x)}{h_k} \phi(x) = \lim_{h_k \rightarrow 0} \int \frac{\phi(x-h_k)-\phi(x)}{h_k} f(x) = -\int \partial \phi(x) f(x) = -<f,\partial \phi>$$

Hence $u$ is the weak derivative of $f$.