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However, the last two terms in (27) can never be zero, and hence we could obtain a stronger inequality.
Finally, we note that when the problem of finding a closed form for the generating function of (Q'_{k}(n)) for arbitrary k will be solved, then further, stronger inequality families will follow by the methods used in this article.
5.3 one can prove a stronger inequality by replacing (varvec{alpha }(E)) with an oscillation index which is the counterpart in our new setting of the one defined in (5.15).
Gaussian distribution (or more general distributions for which the concentration inequality holds [83]), a stronger inequality compared with (20) is valid; this implies that for the reconstruction with a probability of almost one, the following condition for the number of samples m suffices [2, 79]: m ≥ c ′ k log n k (21).
To conclude the proof of the lemma we show that if this is the case, then the following stronger inequality holds begin{aligned} Vert nabla _tau uVert _{infty } le 2 sqrt{Vert uVert _{infty }} frac{1+Vert uVert _{infty }}{1-Vert uVert _{infty }}.
Remark 1 A stronger inequality than (2.4) was given in Lin [[7], Lemma 2.2]: Let A > 0 and any Hermitian B. Then A ♯ ( B A − 1 B ) ≥ B. In what follows, we give the proofs of Theorem 1 and Theorem 2.
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Some other interesting problems concerning stronger inequalities of power exponential functions can be found in [2].
We give a condition which ensures that if one inequality of Sobolev Poincaré type is valid then other stronger inequalities of a similar type also hold, including weighted versions.
Studies which have examined short-term AMI case fatality have frequently shown stronger inequalities in the young and in particular women[ 13, 32- 35].
It follows from Theorem 2.1 that the strong inequality (6) and the strong functional differential inequality (5) for almost all t ∈ I[x] imply the strong inequality (7).
Lemma 3.3 shows that the two-weight strong inequality still holds for differential forms.
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Justyna Jupowicz-Kozak
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