Suppose that: (i) is upper semicontinuous, that is, and with and, implies, (ii), (iii)for each,.
According to Lemma 4, the correspondence D F i is upper semicontinuous.
Hence, Φ i is upper semicontinuous in K with nonempty compact values.
(III) We will prove that Φ i is upper semicontinuous in K with nonempty compact values. .
We will prove that Φ i is upper semicontinuous in K with nonempty compact values.
Since each u i is upper semi-continuous, k ∈ E ⊂ F ( x ).
This shows that the hypograph of mapping T i is closed, hence that T i is upper semi-continuous.
For a given user i, N i is upper bounded by N b i u = R b u i R ( CQI ) l. (4).
Since S i is upper semicontinuous with nonempty and closed values, by Lemma 1.1(i), we find that S i is closed.
Suppose that: (i) is upper semicontinuous, that is,, and, with, and implies, (ii), (iii)for each, there is such that and.
Suppose that: (i) is upper semicontinuous, (ii) satisfies the monotonic condition; for any with and any, there exists such that, (iii there exists an such that.
