Gradient of weak solutions of elliptic PDE at interfaces Let $\Omega$ be a $C^1$-domain partitioned into two open sets $\Omega_1$ and $\Omega_2$ with interface $I=\partial\Omega_1 \cap \partial\Omega_2$.
Now, say $u \in H^1(\Omega)$ is a weak solution of a second order elliptic PDE with piecewise smooth (and bounded) coefficients in $C^\infty(\bar\Omega_1) \cap C^\infty(\bar\Omega_2)$  (that are discontinuous across $I$).
One has $u \in C^\infty(\Omega_1) \cap C^\infty(\Omega_2)$ due to interior regularity. Also, from De Giorgi-Nash-Moser theorem one gets that $u \in C^\alpha(\Omega)$ for some $\alpha >0$ (and thus continous across $I$).
Now, the gradient must be discontinous across $I$ (due to transmission conditions). In this setting, is possible to get any smoothness for the trace of $\nabla u$ from either side (for smooth enough $I$) or can it be arbitrarily bad? From $u \in H^1(\Omega)$, one only knows the trace is in $H^{-1/2}(I)$, so it does not even have to be a function. 
The derivations of transmission conditions for weak solutions (for example here, at page 240) I've seen so far seem to assume that $\nabla u|_{\partial\Omega_1}$ and $\nabla u|_{\partial\Omega_2}$ can be continuously extended to $I$ ad hoc.
 A: It can be arbitrarily bad, I don't see a possibility of a continuous extension unless the singular part is manually removed.
For example, let $\Omega$ be the unit disk in $\mathbb{R}^2$: $\{(x,y): x^2 + y^2 < 1/e\}$, and 
$$u = \ln \Big(\big|\ln (x^2+y^2)\big|\Big).$$
It is easy to check that $u\in H^1_0(\Omega)$ under the polar coordinate system. We have:
$$\nabla u = \left(\frac{2x}{(x^2+y^2)\ln (x^2+y^2)}, \frac{2y}{(x^2+y^2)\ln (x^2+y^2)}\right),$$
Let $\Omega_1$ be the upper disk and $\Omega_2$ be the lower one, so that $I = \{(x,y): x\in (-1/\sqrt{e},1/\sqrt{e}), y=0\}$. The tangential derivative of $u$ on $I$ is:
$$
\frac{\partial u}{\partial s}\Bigg|_I = \nabla u \cdot (1,0)\Big|_I = \frac{1}{x\ln x},\tag{1}
$$
This thing is not even $L^1$-integrable.
Now for the transmission condition, for above $u$, we can see that on $I$ the normal derivative is 0:
$$
\frac{\partial u}{\partial n}\Bigg|_I = \nabla u \cdot (0,1)\Big|_I = 0.
$$
Thus we can easily use this function as a building block to cook a function satisfying the transmission condition, yet still the tangential trace of $u$ on I is not something we can control or continuously extend. For example, letting 
$$
f = \begin{cases}
-\alpha_1\Delta \ln \Big(\big|\ln (x^2+\alpha_1 y^2)\big|\Big) &\text{ in } \Omega_1,
\\[2mm]
-\alpha_2\Delta \ln \Big(\big|\ln (x^2+\alpha_2 y^2)\big|\Big) &\text{ in } \Omega_2,
\end{cases}
$$
where $\alpha = \alpha_1$ in $\Omega_1$ and $\alpha=\alpha_2$ in $\Omega_2$ is a piecewisely defined constant. This $f$ can be proved to be in $H^{-1}(\Omega)$ because it is the Laplacian of a piecewisely defined $H^1(\Omega)$ function $u$, which is the weak solution to $-\operatorname{div}(\alpha \nabla u) = f$. The tangential trace of $u$ will be as bad as (1).

Lastly, I believe you meant to say the page 240 on that book, also notice the obtaining of the transmission condition is equation (6.77), this says that 
$$
\alpha\frac{\partial u}{\partial n} = \beta \frac{\partial u}{\partial n} \; \text{ on } I
\; \text{ in the distributional sense}, 
$$
for we have a smooth testing function with an integration on their product. This only tells us that they agree with each other on $I$ weakly, and tells us nothing about the trace regularity for $\nabla u$. $H^{-1/2}(I)$ is as good as you can get.
