Let $T$ be a one parameter semigroup on Banach space $X$. We know that $T$ has the property $$T(t+s)=T(t)T(s)\quad\text{for all}\quad t,s\ge 0. $$
I was reading some notes on evolution equations and I was told that the above property implies that
- $T(t)=T\left(\frac{t}{n+1}\right)^{n+1}$ and
- $\|T(t)\|\le \|T\left(\frac{t}{n+1}\right)\|^{n+1}$
Now my questions are, first, I do not understand the stated property leads to 1. Next, I do not see how 1. leads to 2.
Appreciate for any help
First, we prove the fact $T(t)=T\left(\frac{t}{n+1}\right)^{n+1}$ by induction. First, we show the property hold for the case $n=1$, i.e. $T(t)=T\left(\frac{t}{2}\right)^{2}$. The right hand side is \begin{equation*} T\left(\frac{t}{2}\right)T\left(\frac{t}{2}\right)=T(t) \end{equation*} equals the left hand side.
Now assume it is true for $n=k$, where $k\ge 2$, we want to show it logically follows that $n=k+1$ is true. So again start with the right hand side we have \begin{align*} \underbrace{T\left(\frac{t}{k+1}\right)\cdots T\left(\frac{t}{k+1}\right)}_{k} T\left(\frac{t}{k+1}\right) &= T\left(\frac{t}{k+1}\right)^k T\left(\frac{t}{k+1}\right)\\ &=T\left(\frac{t}{k+1}\right)^{k+1}=T(t) \end{align*} and so the result 1 follows.
Now using the fact $\|AB\|\le \|A\|\|B\|$, we have \begin{align*} &\ \left\|\underbrace{T\left(\frac{t}{k+1}\right)\cdots T\left(\frac{t}{k+1}\right)}_{k+1}\right\|\\ &\ \le\underbrace{ \left\|T\left(\frac{t}{k+1}\right)\right\|\cdot\left\|T\left(\frac{t}{k+1}\right)\right\|\cdots \left\|T\left(\frac{t}{k+1}\right)\right\|}_{k+1} \end{align*} and so the result 2 follows.