I understand why we must introduce it: If we have some PDE with Dirichlet boundary condition, $$ u|_{\partial\Omega} = g $$
no longer makes sense when $u \in H^1(\Omega)$ since the boundary is a null set. But how does introducing the trace operator actually mend this problem?
Best Answer
The point is that, for $u \in L^2(\Omega)$ for instance, $u\left|_{\partial \Omega} \right.$ cannot be uniquely defined since the boundary has measure $0$ (for $\Omega$ smooth enough at least).
However, it can be defined if $u$ is smooth enough : for instance, if $u$ is continuous, $u\left|_{\partial \Omega} \right.$ is already defined. The trace theorem states that there exists an unique linear and continuous operator (called the trace operator) that extends this trace from continuous functions to $H^1(\Omega)$, with values in $L^2(\partial \Omega)$.
EDIT : You have :
$$ T : C^0(\Omega) \cap H^1(\Omega) \subset H^1(\Omega) \to L^2(\partial \Omega) $$
which is the classical trace operator, well-defined as long as the function is continuous. (Actually, the image is in $C^0(\partial \Omega) \subset L^2(\partial \Omega)$ - at least when $\partial \Omega$ is compact, say.)
Now, you can show that $T$ can be extended uniquely to the whole $H^1(\Omega)$ in a way that preserves linearity and continuity. This means that you can actually give a meaning to the trace of a $H^1$ function - which wasn't obvious at the beginning.
This is particularly useful, as you mentioned it, for partial differential equation : this way, you don't need to ask your solution to be continuous. The space $H^1(\Omega)$ is very well-suited thanks to the trace for equations like $\Delta u = f$, $f$ being a data, and has the advantage of being a Hilbert space (while $C^0$ is not).