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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.
A place belonging to a P-invariant is obviously bounded, i.e. the number of tokens on each place is finite in any reachable marking, and CPI causes structural boundedness, i.e. boundedness for any initial marking.
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These operators are obviously bounded.
Since the integrals over the derivatives of z are obviously bounded due to the estimates of Theorem 3.1, finally we have a look at the last item on the left-hand side of (2.21), κ ( u 2 − z ) ( ψ 2 − ξ ), (5.18).
However, since ( ∫ 0 T ∫ D | q | 2 ( y + δ ) d ( x, y ) d t ) 1 / 2 ≤ c ε, the integrals over items (5.15) and (5.16) can only be estimated by c ε − 1 / 2 τ. (iv) Since the integrals over the derivatives of z are obviously bounded due to the estimates of Theorem 3.1, finally we have a look at the last item on the left-hand side of (2.21), κ ( u 2 − z ) ( ψ 2 − ξ ), (5.18) .
By hypotheses on C, the operator C ˜ is obviously demicontinuous, bounded, and satisfies condition ( S + ).
Bounded sequences are obviously statistically bounded as the empty set has zero natural density.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com