I have a quesition about linear operators on a Banach space.
Let $B$ be a real Banach space. $(T_{t})_{t>0}$ is called strongly continuous contraction semigroup on $B$ if
- For all $t>0$, $D(T_{t})=B$
- For all $t>0$, $\|T_{t}u\|\leq \|u\|$
- For all $t,s>0$, $u \in B$, $T_{t+s}u=T_{t}T_{s}u$
- $\|T_{t}u-u\|\to0$ as $t \searrow 0$
For $\alpha>0$, We can define linear operator $G_{\alpha}:B\to B$ \begin{align*} G_{\alpha}u:=\int_{[0,\infty[}e^{-\alpha t}T_{t}u\,dt\quad({\rm Bochner\,integral}) \end{align*} i.e. $(G_{\alpha})_{\alpha>0}$ is the Laplace transform of $(T_{t})_{t>0}$
My question:
Let $l \in B^{'}(:={\rm all\,linear\,functional\,on }B)$, $u \in B$, $\alpha>0$, \begin{align*} l(G_{\alpha}u)&=\int_{[0,\infty[}e^{-\alpha t}l(T_{t}u)\,dt\\ \int_{[0,\infty[}e^{-\alpha t}l(\alpha G_{\alpha}u)dt&=\int_{[0,\infty[}e^{-\alpha t}l(T_{t}u)dt\quad(int_{[0,\infty[}e^{-\alpha t}dt=\frac{1}{\alpha}) \end{align*} By injectivity of Laplace transform, $l(\alpha G_{\alpha}u)=l(T_{t}u),t>0$. By Hahn Banach theorem, $\alpha G_{\alpha}u=T_{t}u$
Is this an inconsistent argument?
Thank you in advance.
You argument is that $$ F(\alpha) = \int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \int_{0}^{\infty}e^{-\alpha t}\alpha F(\alpha)dt = F(\alpha)=\int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \implies \alpha F(\alpha)=f(t) $$ Does that help you spot your error?