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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!

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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 $(4)$.

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

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