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In this paper, we investigate the commutativity of a prime Banach algebra 𝔄 which admits a nonzero continuous linear generalized derivation 𝔉 : 𝔄 → 𝔄 associated with continuous linear derivation δ such that either 𝔉 ([ 𝔵 m , 𝔶 n ]) − [ 𝔉 (𝔵 m) , 𝔶 n ] ∈ 𝒵 (𝔄) or 𝔉 ([ 𝔵 m , 𝔶 n ]) + [ 𝔉 (𝔵 m) , 𝔶 n ] ∈ 𝒵 (𝔄) for intergers m = m (𝔵 , 𝔶) > 1 and n = n (𝔵 , 𝔶) > 1 and sufficiently many 𝔵 , 𝔶 ∈ 𝔄. Further, similar results are also obtained for unital prime Banach algebra 𝔄 which admits a nonzero continuous linear generalized derivations 𝔉 satisfying either 𝔉 (𝔵 𝔶) n − 𝔉 (𝔵 n) 𝔶 n ∈ 𝒵 (𝔄) or 𝔉 (𝔵 𝔶) n − 𝔶 n 𝔉 (𝔵 n) ∈ 𝒵 (𝔄) for an integer n = n (𝔵 , 𝔶) > 1 and sufficiently many 𝔵 , 𝔶. [ABSTRACT FROM AUTHOR]

The purpose of this paper is to prove the following result which is related to a classical result of Chernoff. Let X be a real or complex Banach space, let ℒ ⁢ (X) {\mathcal{L}(X)} be the algebra of all bounded linear operators of X into itself and let 𝒜 ⁢ (X) ⊂ ℒ ⁢ (X) {\mathcal{A}(X)\subset\mathcal{L}(X)} be a standard operator algebra. Suppose there exist linear mappings ℋ , 𝒢 : 𝒜 ⁢ (𝒳) → ℒ ⁢ (𝒳) {\mathcal{H},\mathcal{G}\colon\mathcal{A(X)}\to\mathcal{L(X)}} satisfying the relations ℋ ⁢ (𝒜 m + n) = ℋ ⁢ (𝒜 m) ⁢ 𝒜 n + 𝒜 m ⁢ 𝒢 ⁢ (𝒜 n) , \displaystyle\mathcal{H}(\mathcal{A}^{m+n})=\mathcal{H}(\mathcal{A}^{m})% \mathcal{A}^{n}+\mathcal{A}^{m}\mathcal{G}(\mathcal{A}^{n}), 𝒢 ⁢ (𝒜 m + n) = 𝒢 ⁢ (𝒜 m) ⁢ 𝒜 n + 𝒜 m ⁢ ℋ ⁢ (𝒜 n) \displaystyle\mathcal{G}(\mathcal{A}^{m+n})=\mathcal{G}(\mathcal{A}^{m})% \mathcal{A}^{n}+\mathcal{A}^{m}\mathcal{H}(\mathcal{A}^{n}) for all 𝒜 ∈ 𝒜 ⁢ (𝒳) {\mathcal{A}\in\mathcal{A(X)}} and some fixed integers m , n ≥ 1 {m,n\geq 1}. Then there exists ℬ ∈ ℒ ⁢ (𝒳) {\mathcal{B}\in\mathcal{L(X)}} , such that ℋ ⁢ (𝒜) = 𝒜 ⁢ ℬ - ℬ ⁢ 𝒜 {\mathcal{H(A)}=\mathcal{AB}-\mathcal{BA}} for all 𝒜 ∈ ℱ ⁢ (𝒳) {\mathcal{A}\in\mathcal{F(X)}} , where ℱ ⁢ (𝒳) {\mathcal{F(X)}} denotes the ideal of all finite rank operators in ℒ ⁢ (X) {\mathcal{L}(X)} , and ℋ ⁢ (𝒜 m) = 𝒜 m ⁢ ℬ - ℬ ⁢ 𝒜 m {\mathcal{H}(\mathcal{A}^{m})=\mathcal{A}^{m}\mathcal{B}-\mathcal{B}\mathcal{A% }^{m}} for all 𝒜 ∈ 𝒜 ⁢ (𝒳) {\mathcal{A}\in\mathcal{A(X)}}. [ABSTRACT FROM AUTHOR]