# When $A \int_0^{\infty} e^{-\lambda t}S(t)u dt = \int_0^{\infty} e^{-\lambda t}S(t)Au dt$?

I have real Banach space $X$ and a bounded linear operator $S: X \to X$ which satisfy:

1) $S(0)u = u$ $\text{ }$ for all $u \in X$

2) $S(t+s)=S(t)S(s)u = S(s)S(t)u$ $\quad$($t,s \geq 0$, $u \in X$)

3) mapping $t \mapsto S(t)u$ $\text{ }$ is continuous from $[0,\infty)$ into $X$

4) $||S(t)|| \leq 1$

I have a closed linear operator $A: D(A) \to X$ where $Au=\lim_{t\to 0+} \frac{S(t)u-u}{t}$ and $D(A)= \{ u \in X; \lim_{t \to 0+} \frac{S(t)u-u}{t} \text{ exists in } X \}$ is dense in $X$.

I have to prove $A \int_0^{\infty} e^{-\lambda t}S(t)u dt = \int_0^{\infty} e^{-\lambda t}S(t)Au dt$.........

I tried: $$I:=\int_0^{\infty}e^{-\lambda t}S(t)udt \quad \text{and} \quad J:=\int_0^{\infty}e^{-\lambda t}S(t)Audt.$$ I have to prove $A(I)=J$. I define $$I_n = \int_0^{n}e^{-\lambda t}S(t)udt \quad \text{and} \quad J_n=\int_0^{n}e^{-\lambda t}S(t)Audt.$$ Therefore $I_n \to I$ and $J_n \to J$. I approximate $I_n$ and $J_n$ with the help of Riemann sums. Divide the interval $[0,n]$ in $k$ parts of length $\frac{n}{k}$. Function $e^{-\lambda t}$ is monotonically decreasing and thus reaches its maximum on the interval $[\frac{i}{n},\frac{i+1}{n}]$ at $\frac{i}{n}$. The approximation with upper Riemann sum for integral $I_n$ is $$I_n = \int_0^{n}e^{-\lambda t}S(t)udt \leq \sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}S(t)udt$$ and for integral $J_n$ is $$J_n = \int_0^{n}e^{-\lambda t}S(t)Audt \leq \sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}S(t)Audt$$ Is this OK?

Then i define sequences $\{ I_{n,k} \}_k = I_{n,1},I_{n,2},\dots$ and $\{ J_{n,k} \}_k = J_{n,1},J_{n,2},\dots$ where $$I_{n,k} = \sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}S(t)udt$$ and $$J_{n,k} = \sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}S(t)Audt$$ Then $I_{n,k} \to I_{n}$ when $k \to \infty$ and $J_{n,k} \to J_{n}$ when $k \to \infty$. Here i also struggle...Is the following true? $$A(I_{n,k}) = A(\sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}S(t)udt) = \sum_{i=1}^k \frac{n}{k} e^{-\lambda(i-1)\frac{n}{k}}\max_{t \in [(i-1)\frac{n}{k},i\frac{n}{k}]}AS(t)udt$$ I know $A$ is linear ($Au=\lim_{t\to 0+} \frac{S(t)u-u}{t}$) but what can i do with the $\max$?

I then use a proven fact in the book, $AS(t)u = S(t)Au$ to get $A(I_{n,k}) = J_{n,k}$. I have $A(I_{n,k}) \to J_n$ and since $A$ is closed i have $A(I_n) = J_n$. Therefore $A(I_n) \to J$. Using again closedness of $A$ i conclude $A(I)=J$.

I would be really glad if someone can read it and help me a bit.

Edit: i think i solved it. In the Riemann sums you can take arbitrary point in the intervals of the partition. You do not need uppper or lower sum. In the limit you get the interval.

• Is it known that $S(s) \int e^{-\alpha \, t} S(t) u \, dt = \int e^{-\alpha \, t} S(s)S(t) u \, dt$...? – saz Apr 27 '13 at 8:07
• Probably since $S(s)$ is continuous operator. – UrošSlovenija Apr 27 '13 at 9:42