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I'm working on Hille-Yosida theorem on Vrabie's book. Here is the statement: statement

In order to prove the sufficiency two lemmas are needed: lemma1

and

lemma2


Here comes my question, is highlighted in yellow:


proof


Why can we deduce that there exists such operator $S(t)$? How can we justify that such limit exist? Because of boundedness? Once we know that there exits is easy to show that $$\|e^{tA_\lambda}x-S(t)x\|\to 0$$ since we have $(3.2.5)$ and $(3.2.3)$.

On the other hand, about the words that are highlighted in blue, the convergence in norm implies uniform convergence, but can we choose that convergence just for compact subsets of $R_+$?.

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1 Answer 1

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In fact, the convergence for any $x\in D(A)$ follows, as said, from (3.2.5), for $0\leqslant a<b$, since we have

$$ \underset{t\in[a,b]}{\sup}\lVert e^{tA_\lambda} x - e^{tA_\mu} x\rVert \leqslant b \lVert A_\lambda x - A_\mu x\rVert \underset{\lambda,\mu \rightarrow \infty}{\longrightarrow} 0.$$

Above limit is true since for all $x\in D(A)$, the following convergence holds in $X$ (see (3.2.3)),

$$ A_\lambda x \underset{\lambda \rightarrow \infty}{\longrightarrow} Ax . $$

Thus, $(e^{tA_\lambda} x)_{\lambda>0}$ is a Cauchy net in $X$ for all $x\in D(A)$, hence admits a limit in $X$ denoted by $S(t)x$.

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