# 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.

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 $$\lim_{t\downarrow 0}\left[\frac{1}{t}(T(t)-T(0))f\right](x) \\ =\lim_{t\downarrow 0}\frac{1}{t}\left[\chi_{[t,\infty)}(x)f(x-t)-f(x)\right] \\ =\lim_{t\downarrow 0}\frac{1}{t}(f(x-t)-f(x))=-f'(x)$$
• Wow very nice! When I was doing this I was splitting a lot of cases, I have lost track after some point. But now I see there is no need for that. It is also nice to see that we get the same formula. I have a question tho, where did we use that $f\in D(A)$? Apparently there is no need to assume that? – Shashi Jun 28 '19 at 9:46
• Oh maybe I see it. What we basically saying in your argument is that if the ILT converges in sup norm then the limit should be, of course, the pointwise limit. We know it converges for $f\in D(A)$ in sup-norm so the claim follows by a density argument. Is this it? – Shashi Jun 28 '19 at 9:54
• @Shashi : The Fourier inversion does require some type of regularity on the function $f$, and differentiability will do it. And, you're right that you can then extend by continuity. This is typical of the Inverse Laplace Transform approach; basically you never get the inversion directly for everything, but you do get it for the domain of $A$ or the domain of $A^2$. The Laplace inversion is not as nice in this regard. The Trotter approximation seems to get around problems more efficiently. But I like the inverse Laplace transform better because it ties in nicely with Heaviside's original work. – DisintegratingByParts Jun 28 '19 at 14:52
• Is Trotter approximation $$T(t) =\lim_{ n\to\infty} \left(\frac n t R(\frac n t, A)\right) ^n$$? I thought that would be unuseful since it would require calculating powers of the resolvent. Is that true? – Shashi Jun 28 '19 at 16:43