I have trouble express explicitly the following $$ \lim_{h\to 0^+} h^{-1}\|f_h-f-hg\|_{L^p(\mathbb R)}=0. $$ which relates to the earlier post in finding generator.
How to find the infinitesimal generator of this semigroup?
Could anyone help writing out the integral?
For the $L^p$ part, I think the following theorem from Hewitt and Stromberg can be viewed as a proof?
This is the translation semigroup (actually, it's a group because you're working on $\mathbb{R}$ and not $[0,\infty)$.) There are various ways of dealing with this problem directly. For example, to show that it is a $C^0$ group, it is enough to show contuity of $T(t)f$ for $f$ in a dense subspace $\mathcal{M}$ of $L^{2}(\mathbb{R})$, such as the subspace $\mathcal{C}_{c}^{\infty}(\mathbb{R})$ consisting of compactly-supported infinitely-differentiable functions. Then, because $\|T(t)f\|=\|f\|$ for all $t$, a simple three $\epsilon$ argument extends the continuity result to all $f\in L^{2}(\mathbb{R})$.
Or, because this group is related to the Fourier transform, you can Fourier transform instead $$ \mathcal{F}(T(t)f) = \mathcal{F}(f(\cdot+t))=e^{its}\hat{f}(s). $$ The Fourier transform turns translation into multiplication by an exponential. From this you immediately obtain the $C^0$ property from the Lebesgue dominated convergence theorem and the Plancherel Theorem: $$ \lim_{t\rightarrow 0}\|T(t)f-f\|^2=\lim_{t\rightarrow 0}\int_{-\infty}^{\infty}|1-e^{its}||\hat{f}(s)|^{2}ds = 0 $$ It is known that $s\hat{f}(s) \in L^2$ iff $f$ is absolutely continuous on $\mathbb{R}$ with $f' \in L^{2}$. For any such $f$, the dominated convergence theorem also gives $$ \lim_{t\downarrow 0}\|\frac{1}{t}(T(t)-I)f-f'\| \\ =\lim_{t\downarrow 0}\|\frac{e^{ist}-1}{t}\hat{f}(s)-is\hat{f}(s)\| \\ = \lim_{t\downarrow 0}\|\frac{1}{t}\int_{0}^{t}(e^{isu}-1)du(is\hat{f}(s))\| = 0. $$ It's a little trickier, but $f$ is in the domain of the generator iff $s\hat{f}(s) \in L^2$.
Added to deal with $L^p(\mathbb{R})$ for $1 \le p < \infty$: The translation group is continuous on every such $L^p$ space, but not on $L^{\infty}$. As mentioned in the first paragraph, the $C^0$ property can be proved by approximating $f\in L^p$ by $g \in \mathcal{C}_c^{\infty}(\mathbb{R})$: $$ \|T(t+h)f-T(t)f\|_p \\ \le \|T(t+h)(f-g)\|_p+\|T(t+h)g-T(t)g\|_p+\|T(t)(g-f)\|_p \\ \le 2\|f-g\|_p+\|T(t+h)g-T(t)g\|_p. $$ Because $g(x+t+h)-g(x+t) \rightarrow 0$ for all $x+t$ as $h\downarrow 0$, then dominated convergence gives $$ \limsup_{h\downarrow 0}\|T(t+h)f-T(t)f\|_p \le 2\|f-g\|_p,\;\;\; g \in \mathcal{C}_c^{\infty}(\mathbb{R}). $$ Because the above holds for all such $g$, then $$ \limsup_{h\downarrow 0}\|T(t+h)f-T(t)f\|_p \le \inf_{g\in\mathcal{C}_c^{\infty}}2\|f-g\|_p = 0. $$ So the $C^0$ property of $T$ holds on $L^p$ for $1 \le p < \infty$.
Now suppose that $f\in\mathcal{D}(A)$, where $A_p$ is the generator of $T$ on $L^p$. Then, by the Lebesgue differentiation theorem, \begin{align} \int_{a}^{b}A_pf\,dt & = \lim_{h\downarrow 0}\frac{1}{h}\int_{a}^{b}(T(h)-I)f\,dt \\ & = \lim_{h\downarrow 0}\frac{1}{h}\int_{a}^{b}f(t+h)-f(t)\,dt \\ & = \lim_{h\downarrow 0}\frac{1}{h}\left[\int_{b}^{b+h}f(t)dt-\int_{a}^{a+h}f(t)dt\right] \\ & = f(b)-f(a) \mbox{ (a.e. a, b)} \end{align} It follows that $f$ can be redefined to be continuous by changing $f$, if necessary, on set of $0$ measure. After doing so, $$ f(b)-f(a) = \int_{a}^{b}A_p f\,dt. $$ Therefore, $f$ becomes absolutely continuous with $$ f'(x) = (A_p f)(x),\;\;\; x\in\mathbb{R}. $$ The converse is similarly proved: If $f$ is absolutely continuous on $\mathbb{R}$ with $f' \in L^p$, then $f \in \mathcal{D}(A_p)$ and $A_pf = f'$.