# Show that the given family of bounded operators on a hilbert space form a semi group.

Suppose $A:D(A)\subset H\rightarrow H$ is a self adjoint, densly defined closed operator and it is also positive operator i.e $<Au,u>\geq0$, for all $u\in H$ ,where $H$ is a hilbert space . Then the operator $-A$ will be a generator of a strongly continuous semigroup ($C_0$ semigroup) and that semigroup will be of the form {$S(t):t\geq0$} , where $S(t):H\rightarrow H$ is given by $S(t)u=$$\int_{0}^{\infty}exp(-\lambda t) dE_\lambda u$ , where {$E_\lambda:\lambda\geq0$} is the spectral resolution of identity correspondenig to the self adjoint positive operator$A$. As $A:D(A)\subset H\rightarrow H$ is positive and self adjoint spectrum of $A$ i.e $\sigma(A)$ is aubset of $(o,\infty)$. Then $(0,\infty)\subset\rho(-A)$ (resolvent set of $-A$).Also $||(-A-\lambda I )^{-1}||\leq$ $\left(\frac{1}{\lambda} \right)$ for all $\lambda >0$. Therefore $-A$ is densly defined closed operator and resolvent set of $-A$ contains $(0,\infty)$ and $||(-A-\lambda I )^{-1}||\leq$ $\left(\frac{1}{\lambda} \right)$ for all $\lambda >0$ . Then by Hille Yosida theorem $-A$ generates a $C_0$ semi group. But I can not able to prove that the above stated family of operators is a semi group. Actually for $t\geq0$ domain of $D(S(t))=${$u\in H: \int_{0}^{\infty}exp(-2\lambda t)d||E_\lambda u||^{2}<\infty$}. Here $\int_{0}^{\infty}exp(-2\lambda t)d||E_\lambda u||^{2}\leq\int_{0}^{\infty}d||E_\lambda u||^{2}=||u||^{2}$ therefore $D(A)=H$ and $||S(t)u||^{2}=\int_{0}^{\infty}exp(-2\lambda t)d||E_\lambda u||^{2}\leq ||u||^{2 }$, $i.e ||S(t)u||\leq||u||$ How do i prove that the stated family of bounded operator is a semigroup? please Help me.

• The spectral integral is a homomorphism. That is, if $f$, $g$ are bounded Borel functions, then $f(A)=\int f(\lambda)dE(\lambda)$ satisfies $(fg)(A)=f(A)g(A)=g(A)f(A)$. – DisintegratingByParts Feb 5 '15 at 20:04
• @T.A.E. thank you. please suggest me some books name for spectral integral to go through the proof of the property that you have used. I don't know spectral theory much. I want to study. – Ajoy Jana Feb 6 '15 at 5:58
• An older basic book with several proofs of the Spectral Theorem and a thorough treatment of spectral resolutions of the identity: "Functional Analysis" Bachman and Narici. amazon.com/Functional-Analysis-Dover-Books-Mathematics/dp/… – DisintegratingByParts Feb 14 '15 at 2:43
• You are welcome. By the way I knew several people who successfully learned on their own from that book. I knew one Professor who had learned from that book, and he commented, "Reading that book is like eating candy." – DisintegratingByParts Feb 18 '15 at 15:38