Does a function that is twice weakly differentiable have a version that is classically differentiable? I have been wondering about the idea of functions that are weakly differentiable. My intuition tells me that the weak derivative allows one to differentiate functions that either have a removable discontinuity or have a kink. Functions with a jump that cannot be "repaired" are not weakly differentiable.   
Furthermore, it seems that if we take the weak derivative of a non-classically differentiable function $f$ with a kink at some point $x$, then the weak derivative will have a jump at $x$ since the limits of the classical derivative approaching $x$ from the left and the right will differ. It would therefore follow that the weak derivative of $f$ is not weakly differentiable. Is this idea correct?  
If so, does that imply that in order for the second weak derivative to exist, the function is equivalent to (in the Lebesgue sense) a classically differentiable function?
 A: Sadly, I did not find any book in which specifically this result is
mentioned, so I will give a proof sketch below.
You might find that you already know some of the facts I use. In this case, skim. Otherwise, I have provided some links where you might find more information on the individual steps.


*

*We start by approximating $f\in W^{1,1}\left(\left(0,1\right)\right)$
(the same works for any open interval of finite length) locally with
smooth functions.


*

*To this end, let us fix any $\varphi\in C_{c}^{\infty}\left(\left(-1,1\right)\right)$
with $0\leq\varphi\leq1$ and $\int_{\mathbb{R}}\varphi\, dx=1$.
Such functions exist, see e.g. here Where can we find examples of smooth functions with compact support? Is there a book?
.


*Now, for $\varepsilon>0$, let $\varphi_{\varepsilon}:=\frac{1}{\varepsilon}\cdot\varphi\left(\frac{x}{\varepsilon}\right)$.
Then $\varphi_{\varepsilon}\in C_{c}^{\infty}\left(\left(-\varepsilon,\varepsilon\right)\right)$,
$\int_{\mathbb{R}}\varphi_{\varepsilon}\, dx=1$ and for every $g\in L^{1}\left(\mathbb{R}\right)$,
we have $\varphi_{\varepsilon}\ast g\xrightarrow[\varepsilon\downarrow0]{L^{1}\left(\mathbb{R}\right)}g$,
where the so-called convolution $F_{\varepsilon}:=\varphi_{\varepsilon}\ast g$
of $\varphi_{\varepsilon}$ and $g$ is given by
$$
\left(\varphi_{\varepsilon}\ast g\right)\left(x\right)=\int_{\mathbb{R}}\varphi_{\varepsilon}\left(x-y\right)g\left(y\right)\, dy.
$$
On convolution in general, see also here Why convolution regularize functions?
and here https://mathoverflow.net/questions/5892/what-is-convolution-intuitively


*

*To show this, we use that $\left\Vert g-L_{x}g\right\Vert _{1}\xrightarrow[x\rightarrow0]{}0$
for every $g\in L^{1}\left(\mathbb{R}\right)$, where $\left(L_{x}g\right)\left(y\right)=g\left(y-x\right)$.
This is shown for example here proof that translation of a function converges to function in $L^1$

*We then have
\begin{eqnarray*}
\left\Vert \left(\varphi_{\varepsilon}\ast g\right)-g\right\Vert _{1} & = & \int_{\mathbb{R}}\left|g\left(x\right)-\int_{\mathbb{R}}\varphi_{\varepsilon}\left(x-y\right)\cdot g\left(y\right)\, dy\right|\, dx\\
 & \overset{\left(\ast\right)}{\leq} & \int_{\mathbb{R}}\int_{\mathbb{R}}\varphi_{\varepsilon}\left(x-y\right)\cdot\left|g\left(x\right)-g\left(y\right)\right|\, dy\, dx\\
 & \overset{\text{Fubini}}{=} & \int_{\mathbb{R}}\int_{\mathbb{R}}\varphi_{\varepsilon}\left(x-y\right)\cdot\left|g\left(x\right)-g\left(y\right)\right|\, dx\, dy\\
 & \overset{z=x-y}{=} & \int_{\mathbb{R}}\int_{\mathbb{R}}\varphi_{\varepsilon}\left(z\right)\cdot\left|g\left(z+y\right)-g\left(y\right)\right|\, dz\, dy\\
 & \overset{\text{Fubini}}{=} & \int_{\mathbb{R}}\varphi_{\varepsilon}\left(z\right)\cdot\left\Vert g-T_{-z}g\right\Vert _{1}\, dz\\
 & \overset{\left(\ast\ast\right)}{\leq} & \sup_{\left|z\right|\leq\varepsilon}\left\Vert g-T_{-z}g\right\Vert _{1}\xrightarrow[\varepsilon\downarrow0]{}0.
\end{eqnarray*}
Here, we used $\int_{\mathbb{R}}\varphi_{\varepsilon}\left(x-y\right)\, dy=1$
at the step marked with $\left(\ast\right)$ and $\int_{\mathbb{R}}\varphi_{\varepsilon}\left(z\right)\, dz=1$
as well as ${\rm supp}\left(\varphi_{\varepsilon}\right)\subset\left(-\varepsilon,\varepsilon\right)$
in the step marked with $\left(\ast\ast\right)$.


*By standard results on differentiation under the integral sign, we
see that $\varphi_{\varepsilon}\ast g$ is $C^{1}$ (actually even
$C^{\infty}$) with derivative
$$
\left(\varphi_{\varepsilon}\ast g\right)'\left(x\right)=\int_{\mathbb{R}}\varphi_{\varepsilon}'\left(x-y\right)\cdot g\left(y\right)\, dy=\left(\left(\varphi_{\varepsilon}'\right)\ast g\right)\left(x\right).
$$
This is for example discussed here Differentiability of Convolutions

*Now comes the interesting part, because we actually want to have $$\left(\varphi_{\varepsilon}\ast f\right)'\left(x\right)=\left(\varphi_{\varepsilon}\ast\left(f'\right)\right)\left(x\right)$$
for $f\in W^{1,1}\left(\left(0,1\right)\right)$. We will show that
this is indeed true for $x\in\left(\varepsilon,1-\varepsilon\right)$.
To see this, note that we have $\gamma_{x,\varepsilon}\in C_{c}^{\infty}\left(\left(0,1\right)\right)$
for 
$$
\gamma_{x,\varepsilon}\left(y\right):=\varphi_{\varepsilon}\left(x-y\right),
$$
as $y\in{\rm supp}\left(\gamma_{x,\varepsilon}\right)$ implies $x-y\in{\rm supp}\left(\varphi_{\varepsilon}\right)\subset\left(-\varepsilon,\varepsilon\right)$
and thus $y\in B_{\varepsilon}\left(x\right)\subset\left(0,1\right)$
by our choice of $x$.
Using the definition of the weak derivative (at $\left(\ast\right)$),
we thus see
\begin{eqnarray*}
\left(\varphi_{\varepsilon}\ast f\right)'\left(x\right) & = & \int_{\mathbb{R}}\varphi_{\varepsilon}'\left(x-y\right)\cdot f\left(y\right)\, dy\\
 & = & -\int_{\mathbb{R}}f\left(y\right)\cdot\gamma_{x,\varepsilon}'\left(y\right)\, dy\\
 & \overset{\left(\ast\right)}{=} & \int_{\mathbb{R}}f'\left(y\right)\cdot\gamma_{x,\varepsilon}\left(y\right)\, dy\\
 & = & \left(\varphi_{\varepsilon}\ast\left(f'\right)\right)\left(x\right).
\end{eqnarray*}

*We know $\varphi_{\varepsilon}\ast\left(f'\right)\xrightarrow[\varepsilon\downarrow0]{L^{1}\left(\mathbb{R}\right)}f'$,
where $f'$ is extended onto all of $\mathbb{R}$ (by zero). Using
$\left(\varphi_{\varepsilon}\ast f\right)'=\varphi_{\varepsilon}\ast\left(f'\right)$
on $\left(\varepsilon,1-\varepsilon\right)$, we get
$$
\left(\varphi_{\varepsilon}\ast f\right)'\xrightarrow[\varepsilon\downarrow0]{L^{1}\left(\left(\delta,1-\delta\right)\right)}f'
$$
for any $\delta\in\left(0,\frac{1}{2}\right)$. But for $x\in\left(\delta,1-\delta\right)$,
we have
$$
F_{\varepsilon}\left(x\right)-F_{\varepsilon}\left(\frac{1}{2}\right)=\int_{1/2}^{x}F_{\varepsilon}'\left(t\right)\, dt=\int_{1/2}^{x}\left(\varphi_{\varepsilon}\ast\left(f'\right)\right)\left(t\right)\, dt\qquad\left(\dagger\right)
$$
and
$$
\left|\int_{1/2}^{x}\left(\varphi_{\varepsilon}\ast\left(f'\right)\right)\left(t\right)\, dt-\int_{1/2}^{x}\left(\varphi_{\theta}\ast\left(f'\right)\right)\left(t\right)\, dt\right|\leq\int_{\delta}^{1-\delta}\left|\left(\varphi_{\varepsilon}\ast f'\right)\left(t\right)-\left(\varphi_{\theta}\ast f'\right)\left(t\right)\right|\, dt\xrightarrow[\varepsilon,\theta\downarrow0]{}0,
$$
where we note that the expression before the limit is independent
of $x\in\left(\delta,1-\delta\right)$.
This shows that the right-hand side of $\left(\dagger\right)$ is
uniformly Cauchy, so that $F_{\varepsilon}\xrightarrow[\varepsilon\downarrow0]{}F$
uniformly on $\left(\delta,1-\delta\right)$ for some (necessarily
continuous) function $F\in C^{0}\left(\left(\delta,1-\delta\right)\right)$.
But we also know $F_{\varepsilon}\xrightarrow[\varepsilon\downarrow0]{L^{1}\left(\left(\delta,1-\delta\right)\right)}f$,
which together implies (why?) $f=F$ almost everywhere.
EDIT: As Harold pointed out in a comment, the uniform Cauchy property of $(\dagger)$ is in general not sufficient to get convergence. But in the current setting, we can fix this as follows: We know $F_\varepsilon \to f$ in $L^1$, so there is some sequence $\varepsilon_n \to 0$ with $F_{\varepsilon_n} \to f$ almost everywhere. Fix some $x_0 \in (\delta, 1-\delta)$ with $F_{\varepsilon_n}(x_0) \to f(x_0)$. We now have $$F_{\varepsilon_{n}}(x)=F_{\varepsilon_{n}}(x)-F_{\varepsilon_{n}}(1/2)-\left(F_{\varepsilon_{n}}(x_{0})-F_{\varepsilon_{n}}(1/2)\right)+F_{\varepsilon_{n}}(x_{0}),$$ where the right hand side converges uniformly by $(\dagger)$ and because $F_{\varepsilon_n}(x_0)$ converges. Now replace all limits $\varepsilon \to 0$ by "going to zero along the sequence $(\varepsilon_n)_n$".

*We conclude that $F$ is a continuous version of $f$ on $\left(\delta,1-\delta\right)$
with
$$
F\left(x\right)=\lim_{\varepsilon\downarrow0}F_{\varepsilon}\left(x\right)=\lim_{\varepsilon\downarrow0}F_{\varepsilon}\left(\frac{1}{2}\right)+\int_{1/2}^{x}F_{\varepsilon}'\left(t\right)\, dt=F\left(\frac{1}{2}\right)+\int_{1/2}^{x}f'\left(t\right)\, dt.
$$
As $\delta\in\left(0,\frac{1}{2}\right)$ was arbitrary, we see
$$
f\left(x\right)=F\left(\frac{1}{2}\right)+\int_{1/2}^{x}f'\left(t\right)\, dt\qquad\left(\ddagger\right)
$$
for almost every $x\in\left(0,1\right)$.

*If $f$ is now two times weakly differentiable, the above argument
shows that $f'$ has a continuous representative, so that the right
hand side of $\left(\ddagger\right)$ is actually a $C^{1}$-function
and thus a $C^{1}$-version of $f$.

