PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 126, Number 1, January 1998, Pages 97{103 S 0002-9939(98)04430-X EXTENDING THE FORMULA TO CALCULATE THE SPECTRAL RADIUS OF AN OPERATOR FERNANDO GARIBAY BONALES AND RIGOBERTO VERA MENDOZA (Communicated by Dale Alspach) Abstract. In a Banach space, Gelfand's formula is used to find the spectral radius of a continuous linear operator. In this paper, we show another way to find the spectral radius of a bounded linear operator in a complete topological linear space. We also show that Gelfand's formula holds in a more general setting if we generalize the definition of the norm for a bounded linear operator. 1. Introduction and basic definitions In all that follows E stands for a linear vector space over the field C of complex numbers. E[t] will denote a complete locally convex topological vector space, with a Hausdorff topology t,and T:E E will be a linear map. Finally, ϑ(t) will be the filter of all balanced, convex and→ closed t-neighborhoods of zero (in E). Definition 1. The linear operator T : E[t] E[t] is said to be a bounded operator, if there is a neighborhood U ϑ(t) such→ that T (U) is a bounded set. ∈ If in the definition above T (U) is a relatively compact set, then T is said to be a compact operator. Any compact operator is a bounded operator, and any bounded operator is continuous (with the t-topology) (see [5]). We recall that, given any topological linear space E[ω]andS:E[ω] E[ω]a linear operator, the resolvent of S is the set → ρ (S)= ξ C ξI S : E[ω] E[ω] ! ∈ − → n is bijective and has a continuous inverse . o The spectrum of S is defined by σ!(S)=C ρ!(S) (the set-theoretic complement in C of the resolvent set), and the spectral radius\ by sr (S)=sup λ λ σ (S) . ! | | ∈ ! n o Definition 2. Anet xα J E is said to be t-ultimately bounded (t-ub) if, { } ⊂ given any V ϑ(t) , there is a positive real number r andanindex α0 J, both depending on∈V , such that x rV α α . ∈ α ∈ ∀ ≥ 0 Received by the editors February 27, 1996. 1991 Mathematics Subject Classification. Primary 47A10; Secondary 46A03. Key words and phrases. Spectral radius, bounded linear operator, locally convex topological space, t-ultimately bounded net. Research supported by the Coordinaci´on de Investigaci´on Cient´ıfica de la UMSNH. c 1998 American Mathematical Society 97 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 98 F. GARIBAY BONALES AND R. VERA MENDOZA Let us denote by Γ the set of all t-ub nets in E. Remark 1. Any bounded, convergent or Cauchy net is a t-ub net. For more details about t-ub nets we refer the reader to [1]. Γ Definition 3. Let ξ C . We will say that 1 T n t 0(Tn=T T ... T , n ∈ ξn −→ ◦ ◦ ◦ times) if, given both V ϑ(t)and xα J Γ , there exist α0 J and n0 N such that 1 T n(x ) V∈ α α {and} ∈n n . ∈ ∈ ξn α ∈ ∀ ≥ 0 ∀ ≥ 0 Γ Definition 4. γ (T ) = inf ξ 1 T n t 0 . t | | ξn −→ n o Remark 2. It is shown by Vera [4] that for a bounded operator T, we have: Γ (i) γ (T ) < , and for any ξ C such that γ (T ) < ξ , 1 T n t 0 . t ∞ ∈ t | | ξn −→ (ii) sr (T ) γ (T ), where sr (T) is the spectral radius of T . t ≤ t t (iii) When E[t] is a Banach space, γt(T )=rt(T). In [2] it was proved, based in the above result, that γt(T )=srt(T )whenTis a compact operator. In this paper we extend that result to any bounded operator. 2. Main results From now on let T : E[t] E[t] be a bounded operator and let U ϑ(t)be such that T (U) is a bounded→ set. ∈ Let PU be the functional of Minkowski (see [3]) generated by U, which is a seminorm on E.LetE[PU] denote the vector space E with the topology given by the seminorm PU . Remark 3. The topology on E given by the seminorm PU is coarser than the topology t (P t ). U ≤ Proposition 1. T : E[PU ] E[PU ] is a bounded operator (hence a continuous one). → Proof. Since T (U) is a bounded set and P t, T(U)isalsoa P -bounded set U ≤ U in E[PU ]. ΓP Definition 5. γ (T ) = inf ξ 1 T n U 0 . PU | | ξn −→ n o Here ΓPU convergence means that, given any net xα J E such that for { } ⊂ all α , PU (xα) r for some positive real number r ( PU -bounded net), then P ( 1 T nx ) ≤0asanetinRwhose index set is N J . U ξn α → × Proposition 2. γPU (T )=γt(T). Proof. Let ξ C be such that γPU (T) < ξ , and let V ϑ(t)and xαJ Γ ∈ 1 | | ∈ { } ∈ be given. Since ξ T (U) is a bounded set, there is a positive real number r1 such 1 that T (U) V . In [1] is shown that xα J Γ r1xα J Γ. This implies r1ξ ⊂ { } ∈ ⇒{ } ∈ that there exist both α J and r > 0 such that r x r U α α , 0 ∈ 2 1 α ∈ 2 ∀ ≥ 0 i.e., PU (r1xα) r2 , that is, the net xα α α is a PU -bounded net; therefore, ≤ { } ≥ 0 α J (α α )andn Nsuch that P ( 1 T n(x )) < 1 α α ,n ∃ 1 ∈ 1 ≥ 0 1∈ U ξn α ∀ ≥ 1 ≥ n , that is, 1 T n(x ) U for those indices. Hence 1 ξn α ∈ 1 1 1 1 n+1 = ( n ) ( ) n+1 T xα T n T r1xα T U V α α1 ,n n1, ξ r1ξ ξ ∈ r1ξ ⊂ ∀ ≥ ≥ License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use SPECTRAL RADIUS OF AN OPERATOR 99 Γ that is, 1 T n t 0 , and therefore, γ (T ) ξ. Thisimpliesthatγ (T) ξn −→ t ≤|| PU ≤ γt(T). On the other hand, let γt(T ) < ξ and xα J , a PU -bounded net; that is, x rU for all α and some r>| | 0. Then{ } 1Tx rT(U), where α ∈ {ξ α}J ⊂ ξ Γ rT(U)isat-bounded set; therefore, 1 Tx Γ . Since 1 T n t 0,given ξ { ξ α}J ∈ ξn −→ >0, α Jand n N such that 1 T n+1x = 1 T n( 1 Tx ) U ∃ 0 ∈ 0 ∈ ξn+1 α ξn ξ α ∈ α α ,n n;thatis,P ( 1 T n+1x ) for those indices. This says ∀ ≥ 0 ≥ 0 U ξn+1 α ≤ that 1 T nx is P -convergent to 0; therefore, γ (T ) ξ . This implies that ξn α U PU ≤| | γ (T ) γ (T ) . t ≤ PU Definition 6. L(E)= S : E[t] E[t] Sis a linear and continuous operator , → n o L (E)= S L (E) S(U) is a bounded set , U ∈ n o L (E) is a vector subspace of the complex vector space L(E). U Remark 4. For the bounded operator T that we have been working on we have n n T, T ,λT,λT LU(E) for all n N and all λ C . Moreover, for any∈ S L(E), S ∈T,T S L (E∈). ∈ ◦ ◦ ∈ U Definition 7. For any operator S LU (E) , we define, taking into account that S(U) is a bounded set, the following∈ real number: S =sup P (Sx) x U . || ||U { U | ∈ } It easy to check that Sn S n S L (E)and n N. || ||U ≤|| ||U ∀ ∈ U ∀ ∈ Γ Theorem 1. If S t S in L(E), then S T S T 0 . n −→ || n ◦ − ◦ ||U → Proof. Let us prove it by way of contradiction. Let >0 be such that there exist natural numbers n1 <n2 <n3 < ... such that < S T S T ; hence, for each of those n there is x U such that || nk ◦ − ◦ ||U k nk ∈ PU [(Snk T S T )xnk )] >.Since Txnk T(U) , it is a bounded sequence; hence, for◦ V−= U◦ ϑ(t) there is an index{ m}⊂ N such that (S S)(Tx ) ∈ 0 ∈ n − nk ∈ V for all n, nk m0 ; this implies that PU [(Snk T S T )xnk )] ,which yields a contradiction.≥ ◦ − ◦ ≤ Proposition 3. ρ (T ) ρ (T ) . t ⊂ PU Proof. Let us suppose first that γt(T ) < 1. Let ξ ρt(T) be such that ξ > 1 k ∈ | | 1 γ (T ). Then S = ∞ k+1 T is a continuous operator and S =(ξI T)− . t k=0 ξ − n 1 k Γt Set Sn = k+1PT .ThenSn S , and from Theorem 1 it follows that k=0 ξ −→ S 1 T S 1 T 0. On the other hand, S 1 T = S 1 I and S || n ◦ ξ −P ◦ ξ ||U → n ◦ ξ n+1 − ξ ◦ 1 T = S 1 I ; hence S S 0 . Thereby, given x E such that ξ − ξ || n+1 − ||U → { m}N ⊂ P (x ) 0,then P (Sx ) P [(S S )x ]+P (S x ) 0.Thisproves U m → U m ≤ U − n m U n m → that S : E[PU ] E[PU ] is a continuous operator; hence ξ ρPU (T ). Now let ξ →ρ (T ) be such that ξ γ(T).