A Relaxed transport
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چکیده
diag(u)Kdiag(v). We have T ⇤1 = diag(u)Kv ) (T ⇤1) a +1 ↵ h(x) a = Kv where we substituted the expression for u. Re-writing T ⇤1, (diag(u)Kv) a +1 = diag(h(x) a )Kv )u a +1 = h(x) a (Kv) a )u = h(x) a a +1 (Kv) a a +1 . A symmetric argument shows that v = y b b +1 (K>u) b b +1 . B Statistical Learning Bounds We establish the proof of Theorem 5.1 in this section. For simpler notation, for a sequence S = ((x 1 , y 1 ), . . . , (x N , y N )) of i.i.d. training samples, we denote the empirical risk ˆ R S and risk R as
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