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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!

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Use the value at $n=0$ in the supremum.

You really want $\leq$ not $<$, since it fails disastrously when $U=0$.

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  • $\begingroup$ Thank you! It helped me and yes I mistyped the inequality. $\endgroup$ – Evan William Chandra Sep 25 '18 at 0:35

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