Is every bounded operator part of a $C_0$-semigroup? Let $X$ be a Banach space and $B \in \mathcal{B}(X)$ be a bounded linear operator on $X$. Is there necessarily a $C_0$-semigroup $T$ such that $B = T(t)$ for some $t$? There might be something obvious I'm missing, but I'm not sure of a good way to approach this problem. The most obvious idea to me would be using some sort of functional calculus for bounded operators that lets you apply a logarithm, and would hopefully result in a (not necessarily bounded) generator for the desired semigroup. I am not aware of any such functional calculus though. I also can't think of a trivial "natural exponential progression" from the identity map to $B$. As far as counter examples go, I know of few theorems that force specific behaviors of $C_0$-semigroups. An obvious one to try is the $0$ operator. At least on $X = C_0[0,1)$, though, the translation semigroup is nilpotent. This is not a homework problem or anything, just something I got curious about.
 A: If $A$ is the generator of a bounded $C_0$ semigroup $T(t)$, then the spectrum of $A$ must lie in the closed left-hand plane $\Re\lambda > 0$ because the resolvent of $A$ is given by the following for $\Re\lambda > 0$:
$$
            \int_0^{\infty}e^{-\lambda t}T(t)xdt=\int_0^{\infty}e^{-\lambda t}e^{tA}xdt=e^{t(A-\lambda I)}(A-\lambda I)^{-1}x|_{t=0}^{\infty}=(\lambda I-A)^{-1}x.
$$
Obviously this is a heuristic argument, but the result is true for all $\Re\lambda >0$. Furthermore, if $M$ is a uniform norm bound for $T$, then
$$
               \|(\lambda I-A)^{-1}\| \le \int_0^{\infty}e^{-\Re\lambda t}dtM\|x\|=\frac{M}{\Re\lambda}\|x\|,\;\;\; \Re\lambda > 0.
$$
So the generator of a bounded $C_0$ semigroup has a resolvent estimate that does not hold for general operators. This type of estimate precludes having a generator $A$ that is nilpotent, for example. To see why, suppose $A^n=0$ for some $n > 1$. Then the following would fail to satisfy the required estimate given above:
$$
           (\lambda I-A)^{-1}=\frac{1}{\lambda}(I-\frac{1}{\lambda}A)^{-1}=\frac{1}{\lambda}\left(I+\frac{1}{\lambda}A + \frac{1}{\lambda^2} A^2 + \cdots+\frac{1}{\lambda^{n-1}}A^{n-1}\right).
$$
By the same token, $A$ cannot have any vector $x\ne 0$ in the domain of $A$ for which $A^nx=0$. So, while the resolvent estimate for $A$ may seem innocuous, it is not. Nilpotent vectors generally keep operators from having all positive roots; however, if $\lambda > 0$, then all positive powers of $(\lambda I-A)$ are defined for generators of a $C_0$ semigroup through the functional calculus. To see why, note that
$$
           \int_0^{\infty}t^{-1+r}e^{-t}dt = \Gamma(r),\;\; r > 0.
$$
By a change of variable for $s > 0$,
$$
    \int_0^{\infty}(st)^{-1+r}e^{-st}d(st)
          = s^{r}\int_0^{\infty}u^{-1+r}e^{-u}du = s^{r}\Gamma(r)
$$
In this way, one may define the following at least on a dense domain:
$$
                A^r = \frac{1}{\Gamma(r)}\int_0^{\infty}u^{-1+r}e^{-u}T(u)du
$$
You can't do this with a general nilpotent operator $A$. Generators of $C_0$ semigroups can be used with a functional calculus that is derived from the Laplace transform, and is related to time evolution systems, which is what the transform was invented for in the first place.
A: A simple counterexample:
A very easy way to get a counterexample is to consider any finite dimensional space $X \not= \{0\}$. Then every $C_0$-semigroup is given by a matrix exponential function, and thus, every operator that occurs in a $C_0$-semigroup is necessarily invertible.
So just take $B$ to be any non-invertible operator on a finite dimensional space.
A general reference on the question:
This article (link to arXiv) by Tanja Eisner deals precisely with the question when a given operator can be embedded into a $C_0$-semigroup.
