Find the semigroup of $Au=-u'$ on the space $C([0,1])$. I have the following problem:

Problem. Consider the Banach space $(C([0,1]),\|\cdot\|_\infty)$. Define the linear operator $(A,D(A))$ by $$D(A)=\{u\in C^1([0,1])\ :\ u'(0)=0\}, \ \ \ Au=-u'$$
  Does $(A,D(A))$ generate a strongly continuous semigroup? Find the semigroup if it does. 


Attemps.
Existence of semigroup. To answer the first question we apply Hille-Yosida for contractions. First of all it is easy to see that $D(A)$ is dense. By solving a simple ODE we see that for $\lambda>0$ we have $\lambda\in \rho(A)$ and
$$R(\lambda,A)f(x)=\frac{f(0)}{\lambda} e^{-\lambda x}+e^{-\lambda x}\int^x_0 f(s)e^{\lambda s}\,ds$$
Note for every $x$
$$|R(\lambda,A)f(x)|\leq \frac{e^{-\lambda x}}\lambda \left| f(0)+\lambda\int^x_0 f(s)e^{\lambda s}\,ds\right|\leq \frac{e^{-\lambda x}}\lambda \left| f(0)\right|+\lambda\int^x_0 \left|f(s)\right|e^{\lambda s}\,ds$$
Hence
$$|R(\lambda,A)f(x)|\leq \|f\|_\infty\frac{e^{-\lambda x}}\lambda \left( 1+\lambda\int^x_0 e^{\lambda s}\,ds\right)=\|f\|_\infty\frac{e^{-\lambda x}}\lambda e^{\lambda x}=\frac{\|f\|_\infty}{\lambda}$$
So Hille-Yosida is applicable and we get the existence of the semigroup of contractions, call them $(T(t))_{t\geq 0}$.
Finding the semigroup.
I know some methods to find the semigroup. For instance I know we can use Yosida-approximations (YA) or Inverse Laplace Transform (ILT) of $R(\lambda,A)$. That is for every $f\in D(A)$
$$\tag{YA} T(t)f=\lim_{\lambda\to\infty}\exp(\lambda AR(\lambda,A)t)f$$
Or ILT is then for every $f\in D(A)$, $\varepsilon>0$ we have
$$\tag{ILT} T(t)f=\lim_{n\to\infty} \frac{1}{2\pi i }\int^{\varepsilon+in}_{\varepsilon-in}e^{\lambda t}R(\lambda,A)f\,d\lambda$$
If I look to both methods, then I see that YA is probably not so nice while ILT seems doable (maybe exchanging integrals at some point is needed there whose justification is hard since the integral does not converge absolutely in general..). 
Anyways, I like those methods since they are generic. However, for this simple $A$ we can also say that $u(t,x)=T(t)f(x)$ is nothing else but the solution of 
\begin{align}
\begin{cases}
u_t=-u_x & \text{ if } (t,x)\in (0,\infty)\times [0,1]\\
u(0,x)=f(x) & \text{ if } x\in [0,1]
\end{cases}
\end{align}
In this case it is pretty easy, it is just a wave-like PDE. The solution of this PDE is $u(t,x)=f(x-t)$, but I'm not sure since $x-t$ can be negative. I think setting $u(t,x)=f(0)$ for $x-t\leq 0$ is pretty reasonable, although I don't know how to justify that. 
So our candidate semigroup is $\tilde T(t)$ defined by
\begin{align}
\tilde T(t)f(x)=
\begin{cases}
f(x-t) &\text{ if } x-t\geq 0\\
f(0) &\text{ if } x-t<0
\end{cases}
\end{align}
Let the generator of this semigroup be $(\tilde A,D(\tilde A))$. We have for $f\in D(\tilde A)$, $x>0$
$$\tilde Af(x)=\lim_{h\to 0^+} \frac{f(x-h)-f(x)}{h}=-f'(x)=Af(x)$$
And for $x=0$ we have
$$\tilde Af(0)=\lim_{h\to 0^+} \frac{f(0)-f(0)}{h}=0$$
These are super useful facts for establishing $(A,D(A))=(\tilde A,D(\tilde A))$, but I don't see how to rigorously finish it.

My questions is actually

Question. How to find the semigroup using PDE representation as I did, or using YA or ILT? I would be interested in all three methods, especially ILT.

Many thanks for your help. 
 A: Your resolvent may be written as
$$
    R(\lambda)f=(A-\lambda I)^{-1}f=\frac{f(0)}{\lambda}+\int_{0}^{x}f(y)e^{-\lambda(x-y)}dy
$$
The inverse Laplace transform method gives
$$
     T(t)f = \lim_{r\rightarrow\infty}\frac{1}{2\pi i}\int_{\epsilon-ir}^{\epsilon+ir}e^{t\lambda}R(\lambda)fd\lambda,\;\;\; \epsilon > 0.
$$
The integral of the first term $f(0)/\lambda$ may be evaluated by closing the contour to the left and evaluating at $\lambda=0$ to obtain $f(0)$. The inversion integral of the second term is
$$
    \lim_{r\rightarrow\infty}\frac{1}{2\pi i}\int_{\epsilon-ir}^{\epsilon+ir}e^{t\lambda}\int_{0}^{x}f(y)e^{-\lambda(x-y)}dyd\lambda \\
  = \lim_{r\rightarrow\infty}\int_0^x f(y)\frac{1}{2\pi i}\int_{\epsilon-ir}^{\epsilon+ir}e^{\lambda(t-(x-y))}d\lambda dy \\
  = \lim_{r\rightarrow\infty}\int_0^x f(y)\frac{1}{2\pi i}\frac{1}{t-(x-y)}(e^{(\epsilon+ir)(t-(x-y))}-e^{(\epsilon-ir)(t-(x-y)})dy \\
  = \lim_{r\rightarrow\infty}\frac{1}{\pi}\int_0^x\frac{\sin(r(t-(x-y))}{t-(x-y)}e^{\epsilon(t-(x-y))}f(y)dy
$$
This is a standard Fourier inversion integral that evaluates to
$$
          e^{\epsilon(t-(x-y))}f(y)|_{y=x-t}=f(x-t),
$$
but only if $y=x-t$ is in the interval $(0,x)$; otherwise it evaluates to $0$. So $(T(t)f)(x)=f(x-t)$ for $0 < t < x$ and is $0$ for $t > x$. This may be written as
$$
                  (T(t)f)(x) = \chi_{[t,\infty)}(x)f(x-t).
$$
Now you can check
$$
         \lim_{t\downarrow 0}\left[\frac{1}{t}(T(t)-T(0))f\right](x) \\
  =\lim_{t\downarrow 0}\frac{1}{t}\left[\chi_{[t,\infty)}(x)f(x-t)-f(x)\right] \\
  =\lim_{t\downarrow 0}\frac{1}{t}(f(x-t)-f(x))=-f'(x)
$$
