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Exact(11)
These operators are obviously bounded.
For any bounded open subset, is obviously bounded.
If ∥ x ′ ∥ p = 0, then ∥ x ′ ∥ p is obviously bounded.
Thus, the node degree of SYG is obviously bounded by k.
To approach this aim, we have to verify that is bounded, where is obviously bounded.
Then { 〈 C x n, x n 〉 } is obviously bounded from above.
Similar(49)
Bounded sequences are obviously statistically bounded as the empty set has zero natural density.
By hypotheses on C, the operator C ˜ is obviously demicontinuous, bounded, and satisfies condition ( S + ).
end{aligned} which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) (mathbf{x}in A), such that (Tmathbf{x}=mathbf{x}).
end{aligned} This implies with the sup norm that |Tx_{1}-Tx_{2}|< |x_{1}-x_{2}|, which shows that T is a contraction mapping on A and therefore there exists a unique solution, obviously a bounded positive solution of (1) (xin A), such that (Tx=x).
Obviously, ({Omega_{3}}) is bounded.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com