The conditions are equivalent in the case that $A$ is bounded. Normally, though, where $A$ may be unbounded with a domain which is not all of $X$, you don't get uniform operator norm convergence because that would imply $A$ is bounded. The strong (vector) convergence is definitely implied by the uniform because $\|Ax\|\le \|A\|\|x\|$ for all $x\in X$ if $A$ is bounded. It's less obvious--but true--that strong convergence implies uniform convergence if the generator $A$ is bounded (this has to do with uniqueness of $C^{0}$ semigroups given the generator, and with the fact that you know how to construct one for a bounded $A$.)
This edit is to flesh out my comment that I made to you. Here's what I was saying you could prove with only minor modification to Pazy's argument:
Theorem: Let $X$ be a Banach space. Let $T : [0,\infty)\rightarrow\mathscr{L}(X)$ be a semigroup. Suppose that there exists $A\in\mathscr{L}(X)$ such that the following limits exist for all $x \in X$:
$$
\lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{X}=0.
$$
Then $T(t)=e^{tA}$. Therefore, for such $T$, one has
$$
\lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{\mathscr{L}(X)}=0,
$$
because the above holds for $e^{tA}$.
Proof: For each $x$, show that $S(t)x=e^{-tA}T(t)x$ has right derivative 0 for all $t \ge 0$. Conclude that $S(t)x=x$ for all $t \ge 0$, which means $e^{-tA}T(t)=I$ for all $t \ge 0$. Multiply on the left by $e^{tA}$ to conclude that $T(t)=e^{tA}$. $\Box$
Here are some basic facts about strongly continuous semigroups that will be useful:
- Let $A$ be the infinitesimal generator of $S(\cdot)$. Then $D(A)$ (the domain of $A$) is dense and $A$ is closed.
- For any $x\in D(A)$ we have $S(t)Ax=\frac{d}{dt}S(t)x$.
- If two strongly continuous semigroups have the same infinitesimal generator, then in fact they are the same semigroup.
Fix $\lambda>0$, $x\in H$ and define $J(t):=\int_0^t S(\tau)x\,d\tau$ ($J$ depends on $x$ as well, but I will omit it for simplicity).
By hypothesis we have $\|J(t)\|\le t\|x\|$, so the integral
$$R(\lambda)x:=\lambda\int_0^\infty e^{-\lambda t}J(t)\,dt$$
makes sense. Let us check that $R(\lambda)=(\lambda I-A)^{-1}$.
$\bullet$ $(S(\epsilon)-I)\int_0^T e^{-\lambda t}J(t)\,dt=\int_0^T e^{-\lambda t}\left(\int_0^t(S(\epsilon)-I)S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left(\int_\epsilon^{t+\epsilon}S(\tau)x\,d\tau-\int_0^t S(\tau)x\,d\tau\right)\,dt$
$=\int_0^T e^{-\lambda t}\left( J(t+\epsilon)-J(\epsilon)-J(t)\right)\,dt$
$=e^{\lambda\epsilon}\int_\epsilon^{T+\epsilon}e^{-\lambda t}J(t)\,dt-\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}J(\epsilon)$
and taking the limit as $T\to\infty$ at the beginning and the end of this chain of equalities and dividing by $\epsilon$ we get
$$\frac{S(\epsilon)-I}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt=\frac{e^{\lambda\epsilon}-1}{\epsilon}\int_0^\infty e^{-\lambda t}J(t)\,dt-\frac{J(\epsilon)}{\lambda\epsilon}-\frac{e^{\lambda\epsilon}}{\epsilon}\int_0^\epsilon e^{-\lambda t}J(t)\,dt$$
But the RHS possesses a limit as $\epsilon\to 0$, namely $R(\lambda)x-\frac{x}{\lambda}$
(the last term tends to $0$ since $e^{-\lambda t}J(t)=o(1)$): thus
$\frac{R(\lambda)x}{\lambda}=\int_0^\infty e^{-\lambda t}J(t)\,dt\in D(A)$ and
$$A\frac{R(\lambda)x}{\lambda}=R(\lambda)x-\frac{x}{\lambda}$$
i.e. $(\lambda I-A)R(\lambda)x=x$.
$\bullet$ Suppose now $x\in D(A)$. The second fact stated at the beginning gives
$\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt
=\int_0^T e^{-\lambda t}(S(t)x-x)\,dt$
$=e^{-\lambda T}J(T)+\lambda\int_0^T e^{-\lambda t}J(t)\,dt-\frac{1-e^{-\lambda T}}{\lambda}x$ (in the last equality we integrated by parts).
Sending $T\to\infty$ we obtain
$$R(\lambda)Ax=\lim_{T\to\infty}\int_0^T e^{-\lambda t}\left(\int_0^t S(\tau)Ax\,d\tau\right)\,dt=R(\lambda)x-\frac{x}{\lambda}$$
so $(\lambda I-A)R(\lambda)x=x$. Moreover $R(\lambda):H\to D(A)$ is a bounded operator.
This proves that $\lambda$ belongs to the resolvent set of $A$ and that $R(\lambda)=(\lambda I-A)^{-1}$.
Finally $\|R(\lambda)x\|\le \lambda\int_0^\infty e^{-\lambda t}\|J(t)\|\,dt
\le \lambda\int_0^\infty e^{-\lambda t}t\|x\|\,dt=\frac{\|x\|}{\lambda}$,
so $\|(\lambda I-A)^{-1}\|\le\frac{1}{\lambda}$.
So Hille-Yosida theorem for contraction semigroups implies that $A$ generates a contraction semigroup, which coincides with $S(\cdot)$.
Best Answer
For this PDE, assuming $c$ is a constant, the explicit solution is given by $u(x,t)=g(x-ct)$ thus your semigroup is merely $S(t)g=g(\cdot-ct)$. You mght use this property to address you problem.