Let $(E, \| \cdot \|)$ be a real Banach space.

Let $A$ and $B$ be two continuous linear operators mapping from $E$ to $E$.

If they have the same operator norm (i.e., $\|A \| = \| B \|$), then could we conclude that their spectral radius are also identical, i.e., $\rho (A) = \rho(B)$?

By Gelfand's formula, we know that $\rho (A) = \lim_{n \to \infty} \| A^n \|^{1/n}$, and $\rho(A) < 1$ implies $\| A^n \| \to 0$ as $n \to \infty$.

It seems that $\|A \| = \| B \|$ may imply $\rho (A) = \rho(B)$ , but I am not sure if it is true. Could someone help me out please?

Thank you very much in advance!

  • $\begingroup$ Wait, does Gelfand's formula apply to real Banach spaces? The proofs I know only apply to complex Banach spaces. $\endgroup$ – Aweygan Apr 16 '18 at 1:25
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    $\begingroup$ I think Gelfand's formula apply to real Banach spaces well, since it also apply to complex Banach spaces. Sorry, I'm quite new in this area and not sure about that:) $\endgroup$ – Paradiesvogel Apr 16 '18 at 1:28

This is not true. Consider the operator $S:\ell^2\to\ell^2$ defined by $$(Px)(n)=\left\{\begin{array}{lcl} x(n)&:& n\text{ even}\\ 0&:& n\text{ odd} \end{array}\right.$$ $S_R:\ell^2\to\ell^2$ the left shift operator, and $I$ the identity operator on $\ell^2$. Put $T=PS_R$. Then $\|T\|=1=\|I\|$, but $T^2=0$, so $\rho(T)=0\neq1=\rho(I)$.


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