Sentence examples for s hence from inspiring English sources

then u ∈ S, hence S is weakly compact in W 1, 2 ( I, R N ).

By Theorem 3.4 F ( F ) = F ( S ), hence x ∈ F ( F ), which completes the proof.

But this state takes the constant value 1/4 on, and hence, sums to 5/4 > 1 on the set S; hence, S is not an event.

Therefore, F is not upper semi-continuous on (C(S)), hence, the graph of F is not closed.

For any sequential process s, Definition 10 ensures that (s in mathrm{sub}(s)); hence, the thesis follows trivially.

By ϱ ∈ Δ 2 s, hence by Lemma 2.2 there exists δ ∈ ( 0, 1 ) independent of x such that ϱ ( x ) > δ.

From Theorem 3.2, we know that operatorname{Fix}(mathcal{F})= operatorname{Fix}(S). Hence (xin operatorname{Fix}(mathcal{F})), which completes the proof of Theorem 4.3.

Since (P2) holds, for (t>0) there exists (ell in mathbb {N}) such that (varphi _{beta }^{ell }(t) le varphi _{beta }(s).) Hence, (varphi _{beta }^{ell +n}(t) le varphi _{beta }^{n}(s)) for all (nin mathbb {N}_0).

By Theorem 2.2(2), S = ∅, hence u > v in Ω and the proof is complete.

Clearly, (bigcap nolimits _{nge 1} V_{(n)}=(0).) It is easy to see that (V_{n,s}subseteqeq V_{ n-s)}.) Hence (V_{ n-seteq V_{(k)}) for ( kge 1,) which implies the assertion of the Lemma.

The function f ∘ h is a sequentially continuous function mapping from S to K ⊆ S. Hence, there exists a fixed point f ∘ h ( Z ) = Z.