Let $A$ be a densely defined closed linear operator in a Banach space $X$ and $\sigma(A)$ be its spectrum. We define its spectral radius $r_{A} := \sup\limits_{\lambda \in \sigma(A)}|\lambda|$. Now, let us fix $r \in \mathbb{R}$ such that $r_{A} < r <\infty$ and define a new norm as follows :
$$ ||U||' = \sup\limits_{n\geq 0} \frac{||A^{n}U||}{r^{n}}, \quad U \in X$$
Claim $||.||'$ is an equivalent norm of $||.||$ in $X$.
I have tried the following :
Claim 1: $\exists C_{1} > 0 \ni ||U||' < C_{1} ||U||$
Proof : We know that $r_{A} = \lim\limits_{n\to \infty} ||A^{n}||^{\frac{1}{n}}$ and $\lim\limits_{n\to\infty}\frac{||A^{n}||}{r^{n}} = 0$ and therefore, we set $C_{1} := \sup\limits_{n\geq 0}\frac{||A^{n}||}{r^{n}}$. Now, observe that
$||U||' \leq \sup\limits_{n\geq 0}\frac{||A||^{n}||U||}{r^{n}} = C_{1}||U||$ and thus we complete the proof for Claim 1.
My problem is to prove the reverse inequality that is $\exists C_{2}>0 \ni C_{2}||U|| < ||U||'$. Any help is very much appreciated! Thank you very much!
Use the value at $n=0$ in the supremum.
You really want $\leq$ not $<$, since it fails disastrously when $U=0$.