the resolvent
Suppose you want to undo an operator A, but A itself has no clean inverse (it is unbounded, it might not even be invertible). The fix is to shift it first: instead of inverting A, invert A minus a number lambda. For most choices of lambda this shifted operator IS nicely invertible, and the inverse is a bounded operator you can actually compute with. That family of shifted inverses, one for each allowed lambda, is the resolvent. It is the single most useful tool for studying an unbounded operator, because it converts hard questions about A into questions about a tame bounded operator.
Precisely: for a closed operator A on a Banach space X and a complex number lambda, the resolvent is R(lambda, A) = (lambda I - A)^{-1}, defined for those lambda where lambda I - A is a bijection from D(A) onto X with bounded inverse. The set of such lambda is the resolvent set; its complement is the spectrum of A (the bad lambda, generalizing eigenvalues). Solving (lambda I - A) u = f means solving the shifted equation lambda u - A u = f, and u = R(lambda, A) f is its unique solution — for the Laplacian this is exactly solving an elliptic equation lambda u - Laplacian u = f, a problem you know how to handle. So the resolvent of -Laplacian is built from solving Helmholtz-type elliptic problems.
Why it is central here: the entire semigroup machinery is phrased through the resolvent. The Hille-Yosida theorem says A generates a C_0 semigroup exactly when its resolvent exists for all large real lambda and satisfies the bound norm of R(lambda, A) <= M/(lambda - omega). The semigroup itself can be recovered from the resolvent by a Laplace-transform-type formula, e^{tA} as a contour integral of e^{t lambda} R(lambda, A). And the location of the spectrum (read off from where the resolvent blows up) tells you the long-time decay rate — a spectral gap below zero forces exponential convergence to equilibrium.
Take A = Laplacian on L^2(Omega) with Dirichlet conditions, lambda > 0. Then R(lambda, A) f = u where lambda u - Laplacian u = f, u = 0 on the boundary — a standard elliptic boundary value problem with a unique solution. As lambda goes to infinity, norm of R(lambda, A) <= 1/lambda, the bound Hille-Yosida wants. The spectrum of A is the set of negative Dirichlet eigenvalues, sitting safely away from the right of the plane.
Inverting (lambda I - A) is solving a shifted elliptic problem.
The resolvent existing for one lambda does not make A invertible — invertibility is the special case lambda = 0, which may fail even when the resolvent set is huge. And the resolvent set is open: the spectrum is what is left over, and for an unbounded operator the spectrum can be unbounded, so 'shift far enough' is exactly why you need lambda large in the generation bounds.