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For each fixed, the function is finite on, because from the definition of the fundamental frequency it follows that (5.36).
It turns out that μ Λ,r,j becomes a translation invariant measure on (mathbb {R}^{d}) which is finite on bounded Borel sets.
Moreover, let ((mathcal{X},d,mu)) be geometrically doubling, (beta>alpha^{n}) with (n=log_{2}N_{0}) and μ a Borel measure on (mathcal{X}) which is finite on bounded sets.
Assume that L is finite on some interval [ 0, c ] with c > 0 and denote by I its Fenchel-Legendre transform on [ 0, c ], I ( x ) = sup 0 ≤ t ≤ c { x t − L ( t ) }.
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All the measures are assumed to be finite on some measure space.
The measures (left (L^{beta }left (cdot,x_{n}right)right)_{nin mathbb {N}}) are finite on (mathbb {R}) and they are supported by [0,τ].
Let N be the set of all counting measures on (mathbb {R}^{d}) which are finite on any bounded Borel set and for which the measure of a point is at most 1.
The interface the hand touches is finite and on it these stresses have to be distributed as uniformly as possible and as low-level as possible so that the human hand doesn't get hurt.
The truth is that time, like money, is finite; spend it on one thing and there's less to go round.
For any q > 0, I q is finite and continuous on ( 0, ∞ ).
Lemma 4.1 (i) For any q > 0, I q is finite and continuous on ( 0, ∞ ).
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com