Smoothness of function 1D sobolev space I came across the following statement in my reading and I need some help to understand it 

If $u \in W^{2,2}(a,b)$, then $u\in C^1([a,b])$

What I know is that if $u\in W^{2,2}(a,b)$, then the first weak derivative of $u'$ is in $L^2(a,b)$ and the second order weak derivative $u''$ is also in $L^2(a,b)$ AND that for a function $f \in W^{1,2}(a,b)$, $f$ is in fact continuous on $[a,b]$.
But, then the only conclusion I can get is that the first weak derivative $u'$ is continuous on $[a,b]$. How do I know that the first classical derivative of $u$ exists?  
 A: You know already that the weak derivative (call it $v$ for the moment) is continuous. By definition, $v$ is called a weak derivative if for all $\phi \in C^\infty_0(a, b)$, we have 
$$\int_a^b \phi' (x) u(x) dx = - \int_a^b \phi(x) v(x) dx.$$
By approximations, the same formula holds when $\phi$ is of compact support and Lipschitz. Now fix any $c, d\in (a, b)$ and let $\phi_n$ be a Lipschitz function defined as 
$$\phi_n (x) = \begin{cases} 1 & \text{if }x \in \left( c + \frac 1n, d - \frac 1n\right)\\ 0 & \text{if } x\notin (c, d) \\ \text{linear} & 
\text{otherwise.}\end{cases}$$ 
Then we have 
$$ n \int_c^{c+ \frac 1n} u(x) dx - n\int_{d-\frac 1n}^d u(x) dx = -\int_a^b \phi(x) v(x) dx.$$ 
Let $n\to \infty$ gives (as suggested by @PhoemueX)
$$u(d) - u(c) = \int_c^d v(x) dx.$$
(Note that the above is true as long as $u\in W^{1, 1}(a, b)$). Now divide by $d-c$ and let $d\to c$, as $v$ is continuous, we have 
$$u'(c) = v(c).$$
Thus we have shown that the derivative of $u$ is really the weak derivative $v$. 
Remark Even when $u$ is only $W^{1, 1} (a, b)$, we can still evaluate the limit
$$\frac{1}{d-c} \int_c^d v(x) dx$$
provided that the limit exist. By the Lebesgue differentiation theorem, this limit exist almost everywhere and is equal to $v(c)$. Thus a $W^{1, 1}(a, b)$ function is differentiable almost everywhere and the derivative is almost everywhere equal to the weak derivative. 
