Difference quotient bounded in $L^1$ does not imply the weak partial derivative exists and is in $L^1$ Let $U\subset \mathbb{R}^{n}$ be open and $V \subset\subset U$ ($V$ is open and compactly contained in $U$). $C$ is a fixed constant.
Given $u\in L^1(U)$. Show by example that if we have $\|D_i^hu\|_{L^1(V)}\leq C$ for all $0<|h|<\mathrm{dist}(V,\partial U),$ it doesn't necessarily follow that the weak derivative $\partial _{x_i}u$ exist. 
Moreover, even if $\partial_{x_i}u$ exists, it may not be in $L^1(V).$ 
Note: $D_i^hu(x):=\frac{u(x+he_i)-u(x)}{h}$
 A: The whole exercise is connected to $L^1$-$L^\infty$ duality and the missing reflexivity of $L^1$.
For the first part, choose e.g. $n=1$, $U=\Bbb{R}$ and $V= (-1,1)$ as well as $u=\chi_{(0,\infty)}$, i.e. the indicator function of the positive reals.
We then have (for $h>0$, the argument for $h<0$ is analogous)
$$
D_1^h u(x)= \begin{cases} 0, & x\leq -h,\\
1/h,& -h < x\leq 0,\\
0,& x>0.\end{cases}
$$
It is then easy to see that this has $L^1$-norm equal to $1$. But $u$ does not have a (locally integrable) weak derivative, because this would imply that $u$ is (locally) absolutely continuous. In particular, it would have a continuous representative (w.r.t. equality a.e.).
For $p\in (1,\infty)$, the reflexivity of $L^p$ prevents this kind of behaviour.
Now for the second claim. Let us assume that $u$ has a weak, locally integrable derivative $\partial_i u$. For $\varphi \in C_c^\infty(V)$, this implies (for $|h| < {\rm dist}(V^c, {\rm supp}(\varphi))$) that
\begin{align}\left|\int \partial_i u \cdot \varphi \, dx \right| &= \bigg|\int u \cdot \partial_i \varphi\, dx\bigg| \\\
&= \left|\lim_{h\to 0} \int u \cdot D_i^h \varphi \, dx\right|\\\
&= \left|\lim_{h\to 0} \int D_i^h u \cdot \varphi \, dx \right|\\\
&\leq C\cdot \Vert \varphi \Vert_\infty\end{align}
The line before that last requires a bit of justification which I leave to you. Hint: Use the facts that Lebesgue measure is translation invariant and that $\varphi$ is compactly supported in $V$.
Using the characterization of the $L^1$-norm using $L^1-L^\infty$ duality, the above implies $\Vert \partial_i u\Vert_1\leq C<\infty$. See also this question $\|g\|_{L^{1}(\mathbb R)}=\sup \{ {|\int_{\mathbb R} fg|: f\in C_{c}^{\infty}(\mathbb R), \|f\|_{L^{\infty}(\mathbb R)}=1\}} ?$.
