Sentence examples for bound of the operator from inspiring English sources

where x ˜ and δ f are defined by (8) and (9), respectively, and m x ˜ is the lower bound of the operator x ˜.

Theorem 5 Let m x and M x, m x ≤ M x, be the bounds of the operator x = ∫ T ϕ t ( x t ) d μ ( t ) and let m x ˜ be the lower bound of the operator x ˜.

Consequently, for the class S consisting of analytic functions f ( z ) = z + ∑ n = 2 ∞ a n z n that are univalent in U, we have the following upper bound of the operator T z α, β.

Furthermore, if f is strictly convex differentiable, then the bound C ¯ 1 K − δ f x ˜ satisfies the following condition: 0 ≤ C ¯ 1 K − δ f x ˜ ≤ { f ( M ) − f ( m ) − f ′ ( m ) ( M − m ) − δ f m x ˜ } 1 K, where m x ˜ is the lower bound of the operator x ˜.

Further, we obtain the upper bound of the operator norm (|A_{k}|_) which implies the continuity of the Hardy-Knopp operator between two different Orlicz spaces.

This means that for actual implementation of CQ algorithm (1.2), one has first to know at least an upper bound of the operator (matrix) norm ∥ A ∥, which is in general difficult.

Instead of estimating the bounds in (1.2), we can try to obtain the upper bound and the lower bound of the norm of the sub-band operator.

We present a method for computing the upper bound and the lower bound of the norm of the sub-band operator based on the theory of circular matrix.

A concept of the sub-band operator of multi-band wavelets is introduced, the theory of d-circular matrices is developed and the upper bound and the lower bound of the norm of the sub-band operator are obtained.

Transcription of the fructose operon is repressed when Cra is bound to the operator, which is located on the downstream region of the RNA polymerase binding site [ 24].

Let λ 0 be the greatest lower bound of spectrum of the operator L 0. Thus the problem − 1 ρ u ″ = λ u + f, u ( 0 ) = 0, u = 0. for λ < λ 0 is uniquely resolvable for any f ∈ L 2 ( R +, ρ ), but it is not if λ = λ 0. Then in view of (A.4) from inequality (4.3) it follows that 1 λ 0 = sup u ≠ 0 ( T u, T u ) [ u, u ] ≤ 4 sup ( s ∫ s ∞ ρ ( x ) d x ).