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Therefore H f ⊂ H ( D ).
Using Lemma 2, we deduce the following inequality: cov h ( cov h ( f, g ) var h ( f ) f − g, cov h ( f, q ) var h ( f ) f − q ) = cov h ( f, g ) cov h ( f, q ) var h ( f ) var h ( f ) cov h ( f, f ) − cov h ( f, g ) cov h ( f, q ) var h ( f ) − cov h ( f, q ) cov h ( f, g ) var h ( f ) + cov h ( g, q ) = cov h ( g, q ) − cov h ( f, g ) cov h ( f, q ) var h ( f ).
However, it satisfies our condition (H f ).
Then we have ξ ⊂ H f (p).
It is clearly, ξ ⊂ H f (p).
Similarly, L h ( f, x ) ≤ f ( x ).
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Lin M-R, Hwang H-F, Yu W-Y, Chen C-Y C-Y
Set h = f g.
Let h ∈ F a b.
for all τ and thus h ≤ f.
Let h = f g be a constant.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com