Motivation of the Proof of the Hille-Yosida Theorem Let $X$ be a Banach space and $A$ be a linear map from a subspace of $X$ to $X$.
The Hille-Yosida theorem gives a necessary and sufficient condition for $A$ to be an infinitesimal generator of a semigroup of class $C_0$.
The precise statement of the theorem is given in the following form.

$A$ generates a semigroup of class $C_0$, say $\left\{ S(t) \right\}_{t\geq0}$ such that $\Vert S(t) \Vert \leq Me^{\omega t}$ with $M>0$ and $\omega\in\mathbb{R}$, if and only if

*

*$A$ is closed and $D(A)$, the domain of $A$, is dense in $X$,

*every real $\lambda>\omega$ belongs to the resolvent set of $A$ and for such $\lambda$ and for all positive integers $n$, $$\Vert (\lambda I - A)^{-n} \Vert \leq \frac{M}{(\lambda-\omega)^n}.$$

To simplify unnecessary complications, let us assume $\omega=0$; there is no loss of generality in assuming this. Moreover, let us denote the map $(\lambda I-A)^{-1}$ simply by $R_{\lambda}$ when $\lambda$ is in the resolvent set of $A$.
The only if part is easy to check.
For the if part, we first construct a family of semigroups, one for each $\lambda>\omega=0$, as follows: Let $A_{\lambda}:=-\lambda I + \lambda^2 R_{\lambda}$ and consider $$S_{\lambda}(t):=e^{tA_{\lambda}}=e^{t(-\lambda I + \lambda^2 R_{\lambda})}=e^{-\lambda t}\sum_{k=0}^{\infty} \frac{\lambda^{2k}t^k}{k!}R_{\lambda}^{k},\quad\text{for each } t\geq0.$$
It can be checked that $\left\{ S_{\lambda}(t) \right\}_{t\geq0}$ is a semigroup of class $C_0$ and satisfies the uniform bound $\Vert S_{\lambda}(t) \Vert \leq M$, by using condition 2 from the statement of the theorem.
It is then proved that for each $u\in D(A)$ and $t\geq0$, the limit $$\lim_{\lambda\to\infty} S_{\lambda}(t)u$$
exists. Subsequently, it is proved that the limit actually exists for all $u\in X$, and we denote this limit by $S(t)u$. After checking subtle convergence and continuity issues, we can show that $\left\{ S(t) \right\}_{t\geq0}$ is a semigroup of class $C_0$ and that $\Vert S(t) \Vert \leq M$ holds. Finally, we show that the infinitesimal generator of this semigroup is exactly $A$, completing the proof.

My question is about how one can come up with the semigroups $\left\{ S_{\lambda}(t) \right\}_{t\geq0}$ in the above proof. What is the motivation for introducing them? I can see that if $\left\{T(t)\right\}_{t\geq0}$ is a semigroup, having $A$ as its infinitesimal generator, we must have
$$T(t)u-S_{\lambda}(t)u=\int_{0}^{t} \frac{d}{ds}[S_{\lambda}(t-s)T(s)u]\,ds=\int_{0}^{t} -S_{\lambda}(t-s)T(s)(A_{\lambda}u-Au)\,ds$$
where $u\in D(A)$ and $t\geq0$. Letting $\lambda\to\infty$ above implies that $T(t)u$ must be the limit of $S_{\lambda}(t)u$, so we see that the semigroup $\left\{T(t)\right\}_{t\geq0}$ is actually unique and that the family of semigroups $\left\{ S_{\lambda}(t) \right\}_{t\geq0}$, indexed by $\lambda$, approximates the semigroup $\left\{T(t)\right\}_{t\geq0}$ as $\lambda$ approaches $\infty$. The argument is very clear but I can't get the idea behind the construction of $\left\{ S_{\lambda}(t) \right\}_{t\geq0}$.
Can someone please explain this, with concrete examples if possible?
 A: The resolvent property was first noticed:
$$
          s-\lim_{\lambda\rightarrow\infty}\frac{\lambda}{\lambda I-A}=I.
$$
That is,
$$
              s-\lim_{\lambda\rightarrow\infty}\lambda R_{\lambda}=I
$$
Essentially, the resolvent has a first order pole at $\infty$ with residue $I$. Operators of this type typically admit some type of spectral expansion because the sum of the residues and integrals around continuous spectrum produce an expansion that will equal the residue at $\infty$, which is $I$. This becomes an expression of the fact that the spectral expansion associated with the finite spectrum is complete. Completeness of the associated spectral expansion is derived from $s-\lim_{\lambda\rightarrow\infty}\lambda R(\lambda)=I$. This approximation to the identity $I$ also can be used to find a bounded approximation to $A$:
\begin{align}
   \lambda R(\lambda)A &=\lambda (\lambda I-A)^{-1}A \\
          &=\lambda(\lambda I-A)^{-1}(A-\lambda I+\lambda I) \\
          &=-\lambda I+\lambda^2R(\lambda).
\end{align}
Using this fact, a bounded approximation to $A$ can be used to generate semigroups. Finally, limits can be used to find the desired semigroup for $A$. The associated bounded approximation is the one you're asking about:
$$
                \exp\{t(-\lambda I +\lambda^2R(\lambda))\}
$$
Hope this explanation helps you see the overall picture. It's a clever technique to define a semigroup for an unbounded generator, provided that completeness is present, which involves proving $s-\lim_{\lambda\rightarrow\infty}\lambda R_{\lambda}=I$. All of this is rooted in Complex Function Theory, and that's why it has a Wizard of Oz feel with the driving force being a residue at $\infty$ equal to $I$. Eigenfunction expansions are generally handled in the same way; at least that's true of the original proofs for the convergence of Fourier type expansions.
