the trace theorem
Here is a real puzzle. A Sobolev function is only defined up to a set of measure zero, and the boundary of a region has measure zero. So 'the value of u on the boundary' ought to be meaningless — you could change u on the entire boundary without changing it as a Sobolev function. Yet boundary conditions are the whole point of a boundary-value problem. The trace theorem rescues the situation by showing that boundary values of an H^1 function DO make sense after all, provided you read them the right way.
The fix is to define boundary values by continuity rather than by pointwise evaluation. Start with smooth functions, where restricting to the boundary is obvious. The trace theorem says this boundary-restriction operation is continuous in the Sobolev norm: if u_n converges to u in H^1, then the boundary restrictions u_n on the boundary also converge, in the boundary L^2 norm. So the operation extends, uniquely and continuously, to all of H^1 — that extension is the trace operator, written T u or u-restricted-to-the-boundary. It is not literal pointwise evaluation; it is the limit you are forced into. Quantitatively, the trace lands in a slightly weaker space than L^2 — the fractional Sobolev space H^{1/2} on the boundary — capturing the precise fact that an interior H^1 function gives up 'half a derivative' on its lower-dimensional edge.
This is what makes the whole Dirichlet story rigorous. The space H^1_0 can now be characterized cleanly as the H^1 functions whose trace is zero — boundary condition and function space line up exactly. Nonzero Dirichlet data g is admissible precisely when it lies in the trace space H^{1/2}, and then there is some H^1 function taking that boundary value (the trace operator is onto H^{1/2}), which you subtract off to reduce to the homogeneous problem. Without the trace theorem, the phrase 'u equals g on the boundary' for a weak solution would be empty; with it, it is a precise statement about an L^2 (indeed H^{1/2}) function on the boundary.
On the unit disc, a function u in H^1 has a trace on the boundary circle that lies in L^2 of the circle. You cannot, however, prescribe that boundary trace to be an arbitrary L^2 function and expect an H^1 extension — only traces in the smaller space H^{1/2} of the circle extend to H^1 functions inside. A wildly discontinuous boundary value (a step around the circle) is in L^2 but NOT in H^{1/2}, and indeed it cannot be the trace of any finite-energy function.
The trace of an H^1 function lives in H^{1/2}, not all of L^2 — half a derivative is lost at the edge.
The trace is NOT pointwise evaluation — for a fixed rough u there is no single 'value at a boundary point', only the L^2/H^{1/2} trace defined by continuity. Also the theorem needs a reasonably regular boundary (Lipschitz is enough); on a fractal or cusped boundary the trace can misbehave. And L^2 functions (with no derivative control, H^0) have NO well-defined trace at all — you need at least the bit of smoothness that H^1 provides.