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.
Best Answer
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:
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: