Infinitesimal generator of a C0-semigroup

For a Banach space $X$, the operator vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores is extended as follows: an operator is a linear map $A:D(A)\to X$ on a linear subspace of $X$, its graph is $\mathrm{\Gamma }(A)=\{(x,Ax):x\in D(A)\}\subseteq X\times X$, it is closed when this graph is closed, and densely defined when $D(A)$ is dense in $X$. Use $\mathrm{\parallel }(x,y)\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }y\mathrm{\parallel }$ on $X\times X$ and the graph norm $\mathrm{\parallel }x{\mathrm{\parallel }}_A=\mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }Ax\mathrm{\parallel }$. These norms are equivalent to the square-sum norms in the Hilbert suppliers. The product is Banach because its two coordinate Cauchy sequences converge in $X$; the closed graph is therefore Banach, and $x\mapsto (x,Ax)$ is an isometry of the graph-norm domain onto it. Under Countable Choice, sequential closedness is equivalent to closedness: for any point in a closure, choose graph points within $1\mathrm{/}n$ and pass to their limit.

Let $(T(t){)}_{t\ge 0}$ be a strongly continuous semigroup on a Banach space $X$ (Strongly continuous semigroup). Its infinitesimal generator is the linear operator $A:D(A)\subseteq X\to X$ with domain $D(A):=\{x\in X:{\lim }_{t\downarrow 0}\frac{T(t)x-x}{t}\text{ exists in }X\}$ and $Ax:={\lim }_{t\downarrow 0}\frac{T(t)x-x}{t}$ for $x\in D(A)$. The limit is a one-sided limit at the boundary point $0$, and $D(A)$ is a linear subspace of $X$ (Normed subspace); the operator is recorded as the pair $(A,D(A))$ in the sense of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores. Neither boundedness nor closedness of $A$, nor density of $D(A)$, is assumed in the definition; under its stated choice hypothesis, The generator is closed and densely defined proves that the graph of $A$ is closed and $D(A)$ is dense in $X$. The domain need not be closed in the norm of $X$.

The one-sided limit. For $x\in D(A)$ the difference quotient $\frac{T(t)x-x}{t}\in X$ is defined for every $t>0$, and the defining limit is taken along $t\downarrow 0$ only; no two-sided limit at the boundary point $0$ of $[0,\mathrm{\infty })$ is considered, and the vector $Ax$ is the limit when it exists. The value $Ax$ is unique because $X$ is a metric space (Banach space).

The domain is a linear subspace. The zero vector lies in $D(A)$ and $A0=0$. If $x,y\in D(A)$ and $\alpha ,\beta$ are scalars, then by linearity of each $T(t)$ (A bounded linear operator between normed spaces) the difference quotient of $\alpha x+\beta y$ equals $\alpha \frac{T(t)x-x}{t}+\beta \frac{T(t)y-y}{t}$ for every $t>0$; as $t\downarrow 0$ this converges to $\alpha Ax+\beta Ay$, because vector addition and scalar multiplication are continuous and scalar multiplication by the fixed scalars $\alpha ,\beta$ is continuous. Hence $\alpha x+\beta y\in D(A)$ with $A(\alpha x+\beta y)=\alpha Ax+\beta Ay$, so $D(A)$ is a linear subspace of $X$ (Normed subspace) and $A$ is linear on it. The operator is recorded as the pair $(A,D(A))$ in the vocabulary of Unbounded linear operators: domain, graph and extension and Densely defined, closed and closable operators, and cores.

What is not assumed. Boundedness and graph closedness of $A$, and density of $D(A)$ in $X$, are not defining assumptions. Graph closedness and domain density are conclusions of the later theorem under its stated choice hypothesis; they do not assert that $D(A)$ is closed in the norm of $X$. The semigroup axioms used here are those of Strongly continuous semigroup, in particular $T(t)$ is everywhere defined and bounded for every $t\ge 0$.