# Questions from An Introduction to Semilinear Evolution Equations Corollary 4.1.9

Before, proceeding to the problem, I will mention some notations on this book to be easier to read.
0. $$(X, ||\, \cdot\,||)$$ is Banach space
1. $$L^{1}((0,T),X)$$ : the space of measurable functions $$u$$ on $$(0,T)$$ with values in $$X$$ such that $$||u||$$ is integrable.
2. $$D'(\Omega)$$ : the space of distributions on $$\Omega$$.
3. $$u' = u_{t} = \frac{du}{dt}$$ for $$u \in D'((0,T),X)$$
4. $$W^{1,1}((0,T),X) := \{u \in L^{1}((0,T),X),u' \in L^{1}((0,T),X), \text{in the sense of } D'((0,T),X)\}$$
5. $$||u||_{L^{1}((0,T),X)} = (\int_{0}^{T}||u||dt)^{1/p}$$
6. $$||u||_{W^{1,1}((0,T),X)} = ||u||_{L^{1}((0,T),X)} + ||u'||_{L^{1}((0,T),X)}$$
7. $$(D(A),||\, \cdot \,||_{D(A)})$$is Banach Space with $$||u||_{D(A)}= ||u|| + ||Au||$$ when $$A$$ is a linear operator with a closed graph.
8. In this problem, $$A$$ is $$m$$-dissipative operator with dense domain and $$(\mathscr{T}(t))_{t\geq 0}$$ is the contraction semigroup generated by $$A$$.
9. $$C([0,T],X)$$ : space of continuous functions from $$[0,T]$$ to $$X$$.
Now, I will state the corollary.

Corollary 4.1.9
Let $$x \in X$$, $$f \in L^{1}((0,T),X)$$, and $$u \in L^{1}((0,T),X)$$. Assume further that $$u \in L^{1}((0,T),D(A))$$ or $$u \in W^{1,1}((0,T),X)$$. Then we have $$u(t) = \mathscr{T}(t)x + \int_{0}^{t}\mathscr{T}(t-s)f(s)ds \tag{1}$$ if and only if $$u$$ satisfies
(i). $$u \in L^{1}((0,T),D(A))\cap W^{1,1}((0,T),X)$$
(ii). $$u'(t) = Au(t) + f(t)$$, for almost every $$t \in [0,T]$$
(iii). $$u(0) = x$$

Before I explain the part where I do not understand, I would like to state some theorems used to prove this corollary.

Theorem (Extrapolation)
Let $$X$$ be Banach space and $$A$$ is $$m$$-dissipative with dense domain. Then, there exists $$\bar{X}$$, and $$\bar{A}$$ such that the Banach Space $$(\bar{X},||\, \cdot\,||_{\bar{X}})$$ satisfying:
(I). $$X \hookrightarrow \bar{X}$$ with dense embedding.
(II). $$\forall x \in X, \, ||x||_{\bar{X}} = ||(I-A)^{-1}x|| \leq ||x||$$
Moreover, $$A$$ satisfies
(I). $$D(\bar{A}) = X$$
(II). $$\forall x \in D(A), \, \bar{A}x = Ax$$
Lemma 4.1.5
$$\forall x \in X, \forall f \in L^{1}((0,T),X)$$, formula (1) defines a function $$u\in C([0,T],X)$$. In addition, we have $$\|u\|_{C([0,T],X)} \leq ||x|| + ||f||_{L^{1}((0,T),X)}$$
Now, I will state the part in which I do not understand in the textbook. So, we assume $$(1)$$ holds. Now, let $$(f_{n})_{n\geq 0}$$ be a sequence of $$C([0,T],X)$$ such that $$f_{n} \to f$$ in $$L^{1}((0,T),X)$$ as $$n\to \infty$$. Let $$u_{n}$$ be the corresponding solutions of (1). From Corollorary 4.7 (I understand the reasoning so I will not mention Corollorary 4.7 here) we will obtain $$\forall t \in [0,T], \, u_{n}'(t) = \bar{A}u_{n}(t) + f_{n}(t)$$ The book claims by using Lemma 4.1.5, we can show that $$u(t) = x + \int_{0}^{t}\bar{A}u(s) + f(s)ds$$ How do we get the result above by using Lemma 4.1.5? I try but to no avail using the standard $$|u_n(t)-u(t)|$$ trick. Any help will be much appreciated!

Recall that $$u_n(t) = \mathscr{T}(t)x + \int_{0}^{t}\mathscr{T}(t-s)f_n(s)ds\tag{1}$$ and thus $$u_n(t)-u(t)=\int_{0}^{t}\mathscr{T}(t-s)(f_n(s)-f(s))ds.$$ Therefore, from Lemma 4.1.5 applied to $$u_n-u$$, $$\|u_n-u\|_{C([0,T],X)} \leq \|0\| + \|f_n-f\|_{L^{1}((0,T),X)}\overset{n\to\infty}{\longrightarrow} 0$$ which implies that $$u_n(t)\overset{n\to\infty}{\longrightarrow} u(t),\quad\forall\ t\in[0,T]\tag{2}.$$

On the other hand, from $$u_{n}'(t) = \bar{A}u_{n}(t) + f_{n}(t)$$ it follows that $$u_{n}(t) =u_n(0)+ \int_0^t\bar{A}u_{n}(s) + f_{n}(s)\;ds \overset{(1)}{=}x+ \int_0^t\bar{A}u_{n}(s)\;ds + \int_0^tf_{n}(s)\;ds\tag{3}$$ which implies (see Remark below) that $$u_n(t)\overset{n\to\infty}{\longrightarrow} x+ \int_0^t\bar{A}u(s) + f(s)\;ds,\quad\forall\ t\in[0,T]\tag{4}.$$

From $$(2)$$ and $$(4)$$ we obtain $$u(t)=x+ \int_0^t\bar{A}u(s) + f(s)\;ds.$$

Remark. It is clear that $$\int_0^t f_{n}(s)\;ds\overset{n\to\infty}{\longrightarrow}\int_0^t f(s)\;ds\tag{5}$$ Since

• $$\bar{A}$$ is closed

• $$\displaystyle\int_0^tu_{n}(s)\;ds\overset{n\to\infty}{\longrightarrow}\int_0^tu(s)\;ds$$ in $$X=D(\bar{A})$$

• $$\displaystyle\bar{A}\int_0^tu_{n}(s)\;ds\overset{(3)}{=}u_{n}(t) -x - \int_0^t f_{n}(s)\;ds$$ is convergent

we conclude that

$$\int_0^t\bar{A}u_{n}(s)\;ds=\bar{A}\int_0^tu_{n}(s)\;ds\overset{n\to\infty}{\longrightarrow}\bar{A}\int_0^tu(s)\;ds=\int_0^t\bar{A}u(s)\;ds.\tag{6}$$

Substituting $$(5)$$ and $$(6)$$ into $$(3)$$ we obtain $$(4)$$.

• Thank you very much for your clear and precise explanation! Now I understand the part I did not understand previously – Evan William Chandra Oct 22 '18 at 2:57