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I try to use the following problem:

Let $A$ be the generator of the $C_0$-semigroup $(T(t))_{t \geq 0}$ on a Banach space $X$. Let $\varphi \in C_c^\infty(0, \infty)$ and $T_\varphi x = \int_0^\infty \varphi(s) T(s) x \, ds$.

Now I want to show that $D = \{T_\varphi x : \varphi \in C_c^\infty(0, \infty), x \in X \}$ is dense in $X$. So I need to construct a sequence of mollifiers $(\varphi_n)_{n \in \mathbb N}$ in $C_c^\infty(0, \infty)$ such that $T_{\varphi_n} x \to x$ in $X$ for $n \to \infty$.

But I really struggle to write down such a suitable sequence that has the desired properties. I guess its must be build such that $\operatorname{supp} \varphi_n$ "runs against" $0$ for $n \to \infty$ while being bounded from above for all $n \in \mathbb N$, because I need a bound for $\Vert T(t) \Vert$ on $\bigcup_{n \in \mathbb N} \operatorname{supp} \varphi_n$.

I would appreciate some hints on the topic. Thanks in advance :)

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Let $\varphi \in C_c^{\infty}((0,\infty))$ be a non-negative function such that $\int_{(0,\infty)} \varphi(x) \, dx = 1$ and the support of $\varphi$ is contained in $(0,1)$. Define

$$\varphi_n(x) := n \varphi \left( nx \right).$$

Using that $\int \varphi_n(x) \,dx=1$ and $\text{spt} \, \varphi_n \subseteq B(0,1/n)$ we find for any $x \in X$,

$$\begin{align*} \|T_{\varphi_n}x-x\| &= \left\| \int_0^{\infty} \varphi_n(s) (T(s)x-x) \, ds \right\| \leq \sup_{s \leq 1/n} \|T(s)x-x\|. \end{align*}$$

Since $(T(t))_{t \geq 0}$ is strongly continuous at $t=0$, the right-hand side converges to $0$ as $n \to \infty$.

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