# Equivalence of norm in Banach Space

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!

## 1 Answer

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

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

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