The first approach is to define μ in the intensity domain (I), i.e., muleft(B x,varepsilon)right) = int_{B x,varepsilon)}{(G_{varepsilon} ast I)} dx, (7).
Another choice is to define μ(B x,ε)) as the sum of the Laplacians of the image inside B x,ε), that is: mu(B x,varepsilon)) = int_{B x,varepsilon)}|nabla^{2} (G_{varepsilon} ast I)| dx.
An independent casuistry comprised of 22 (NN), 23 (AST I), 26 (AST II), 18 (AST III), 83 (AST IV or GBM), 25 (OLI II), and 26 (OLI III) was analyzed at the validation step by qRT-PCR for the selected targets.
The incremental selection AST (I-AST) adds successively antennas at every stage, so the antenna that yields the maximum increase of the channel capacity is added to the set of P transmit antennas [14].
Then, we define a measure function μ(B x,ε)) for the image I as follows: mu(B x,varepsilon)) = left int_{B x,varepsilon)}sumlimits_{k}left(,f_{k} ast (G_{varepsilon} ast I right)^{2} dxright)^{1/2}.
Notice that (pi ^{ast }_{i} > 0) means that state i is not a transient state.
Notice that the condition (pi ^{ast }_{i}>0) indicates that state i is not a transient state.
Let (V_{S}=mathbb{R}mbox operatorname{span}{X_{i}^{ast}: i in S}), (V_{T}=mathbb {R}mbox operatorname{span}{X_{i}^{ast}: i in T}), and dξ be the Lebesgue measure on (V_{T}) such that the unit cube spanned by ({X_{i}^{ast}:i in T}) has volume 1.
(Deterministic optimal solution) The optimal solution of P- 4 can be achieved by a deterministic power allocation policy, i.e., (p^{ast }_{i,j} in {0, 1}).
Because (x^{ast} inmathbb{F}), (t^{m} inhat {T}_{j_{m}}(x^{ast})), (m inunderline{nu_{0}}), and (lambda^{ast}_{i} = varphi_{i}(x^{ast})), (i inunderline {p}), the right-hand side of the above inequality is equal to zero, and hence we have (L x,u^{ast},v^{ast},lambda ^{ast},bar{t},bar{s}) geqq0).
To sum up, with the assumption that for non-transient states i,m, (p^{ast }_{i,j} > 0) and (p^{ast }_{m,n} > 0, m > i, n < j), it ends up that P ∗ is not optimal.
