Sobolev norm in the definition of Sobolev spaces I've seen the Sobolev space defined as:

The Sobolev space $H^k(\Omega)$ is the set of all functions $u \in L_2(\Omega)$ for which the weak derivative $\partial^\alpha u \in L_2(\Omega)$ for all $|\alpha|\leq k$ exist.

But shouldn't also the Sobolev norm exist (be finite)?
For example $u(x) = x^{-1/4}$ on $\Omega = (0,1)$ is in $L_2(\Omega)$ and even its classical derivative (and so its weak derivative) exists but since
$$ \int_\Omega |u'(x)|^2 \, dx = \int_0^1 \frac{1}{16} x^{-5/2} , dx = \biggl[-\frac{1}{24} x^{-3/2} \biggr]_0^1 = \infty$$
the Sobolev norm 
$$ \|u\|_{H^1(\Omega)} = \sqrt{\int_\Omega |u(x)|^2 \, dx + \int_\Omega |u'(x)|^2 \, dx}$$
 can't be finite.
 A: Shouldn't also the Sobolev norm exist (be finite)? 
Yes, but this condition is automatically satisfied because of the way that the space was defined. In fact, by definition
$$L_2(\Omega)=\left \{f:\Omega\to\mathbb{R};\;f\text{ is measurable and }\int_\Omega  |f(x)|^2 \ dx<\infty\right\}.$$
So,  "$u \in L_2(\Omega)$" implies
$$\int_\Omega  |u(x)|^2 \ dx<\infty$$
and "$\partial^\alpha u \in L_2(\Omega)$ for all $|\alpha|\leq k$" implies
$$\int_\Omega  |\partial^\alpha u(x)|^2 \ dx<\infty,\quad \forall\ |\alpha|\leq k.$$
Thus, your definition implies that $\|u\|_{H^k(\Omega)}$ is finite (that is, the "existence" (finiteness) of the Sobolev norm is implicitly stated).
Remark: When you said and so its weak derivative, you were using the following property:

If $u$ possess classical derivative $u'$, then $u$ possess weak derivative and thus $u\in H^1(0,1)$. Moreover, $u'$ coincides with the weak derivative of $u$.

However, by definition of $H^1(0,1)$, the weak derivative of a function in $H^1(0,1)$ have to be a function in $L_2(0,1)$ and thus the above property is false because sometimes $u'$ exists but doesn't belong to $L_2(0,1)$ (that is, doesn't satisfy
$\int_0^1  |u'(x)|^2 \ dx<\infty.$
An example is given by your function $u(x)=x^{-1/4}$). 
The valid property is:

If $u \in C^1(0,1)\cap L_2(0,1)$ and if $u'\in L_2(0,1)$, then $u\in H^1(0,1)$ . Moreover, $u'$ coincides with the weak derivative of $u$.

