Consider the function $u(x) = x^{\frac{1}{2}}$ on the domain $[0, 1]$. This function is an element of $W^{1, 1}$ and $W^{1, \infty}$ but not $W^{1, 2}$ as
- for $W^{1, 1}$, we have $\Vert\frac{\partial u}{\partial x}\Vert_1 = 1 < \infty$.
- for $W^{1, \infty}$, we have $\Vert\frac{\partial u}{\partial x}\Vert_\infty = \frac{1}{2} < \infty$.
- for $W^{1, 2}$, we have $\Vert\frac{\partial u}{\partial x}\Vert_2 = \frac{1}{4}(\ln 1 - \ln 0) = \infty$.
In fact, $u \in W^{1,p}$ only if $p < 2$ or $p = \infty$. What intuition should I take from this situation? If I take $p < 2$ and gradually let it approach I can see that $u$ from evaluating $\Vert\frac{\partial u}{\partial x}\Vert_p$ that $u$ is getting 'closer and closer' to not being an element of $W^{1,p}$. How does someone with experience of Sobolev spaces view this phenomenon?
Your phenomenon cannot be true as we have $W^{1,p}(0,1) \subseteq W^{1,q}(0,1)$ if $1 \leq q < p \leq \infty$. And indeed, for $1 \leq p < \infty$, we have
$$ \int_0^1 u'(x)^p \, dx = \frac{1}{2^p} \int_0^1 x^{-\frac{p}{2}} \, dx = \begin{cases} -\frac{1}{2^p \left( \frac{p}{2} - 1 \right)} \left(1 - \lim_{x \to 0} x^{-\frac{p}{2} + 1} \right) & p \neq 2, \\ -\frac{1}{4} \lim_{x \to 0} \ln(x) & p = 2. \end{cases} $$
Thus, the integral converges if and only $-\frac{p}{2} + 1 > 0$ or $1 \leq p < 2$.
For $p = \infty$, we have
$$ ||u'(x)||_{\infty} = \sup_{x \in [0,1]} \left| \frac{1}{2} x^{-\frac{1}{2}} \right| = \infty $$
which implies that $u \in W^{1,p}$ if and only if $1 \leq p < 2$.