Sentence examples for mu for all from inspiring English sources

Then Ax exists and is in (B(mu)) for all (x=(x_{mn})in B lambda)) and ({B(Ax)}_{mn}in mu) for all (m,ninmathbb{N}).

Then the Córdoba maximal operator (mathbb{M}_{mu}) is bounded on (L^{p} omega,dmu)) if and only if (omegainmathbb{A}_{p}(mu)) for all (1< p

Thus, (A=(a_{mnkl})in(lambda:mu)) if and only if the double series on the right side of (1.1) converges in the sense of ϑ for each (m,ninmathbb{N}), i.e., (A_{mn}inlambda ^{beta vartheta)}) for all (m,ninmathbb{N}) and we have (Axin mu) for all (xinlambda); where (A_{mn}=(a_{mnkl})_{k,linmathbb {N}}) for all (m,ninmathbb{N}).

Therefore, so long as the bias terms' sums converge, the estimates will converge to the true global values, that is ({tilde{h}}_i^{ k)}rightarrow {frac{1}{N}}h(x^{ k)})) and ({tilde{mu }}_i^{ k)}rightarrow mu) for all (iin N), where (mu) is the global estimate of the optimal multiplier (mu ^*).

From (H4) it is obvious that (x^{top}H(mu,x <0) for all ((mu,x in 0,1 times (partialOmegacapoperatorname{Ker} L)).

(Bsubseteq C) implies (mu (B le mu (C),) for all (B,Csubseteq X;)   3.

(Bsubseteq C) implies (mu (B le mu (C),) for all (B,Csubseteq X;).

By inequality (41), we deduce that (mu _nle nu _1), and by Theorem 4, we have (,mu _nge mu _1,) for all (,nin mathbb {N}).

Therefore, there exists an integer k ′′>k ′ such that |I μ (x (k))|≥K and (0le sigma (x^{ k)}))_{n}mu }) for all k≥k ′′.

By the assumption, we take (varrho_{2}>0) such that (Q x geq Q infty)+ xi_{2} vert x vert ^{-2^{astast}(Lambda_{0}-l_{1}(mu))}) for all (vert x vert geqvarrho_{2}).

In our experiments, we set (mu =0.005) for all three methods.