We complete the uniqueness proof.
Therefore, we have and this completes the uniqueness property of and We can prove the other results similarly.
It feels less overwhelming to, for example, put others into a labelled box than to recognize their complete individuality and uniqueness.
Lemma 5 and Lemma 6 yield E ( sup t 0 ≤ s ≤ t | X ( s ) − X ¯ ( s ) | 2 ) = 0, t 0 ≤ t ≤ T. The proof of uniqueness is complete.
If these counterforces become strong enough, you can lose complete contact with your uniqueness, with who you really are.
This contradiction completes the proof of uniqueness.
This completes the proof of uniqueness.
Therefore letting, one has for all, completing the proof of uniqueness.
Therefore one obtains Q ( x ) = Q ′ ( x ) for all x ∈ X, completing the proof of uniqueness.
Since w ( 0, x ) = w x ( 0, x ) = 0, we obtain u ( t, x ) = v ( t, x ) for a.e. ( t, x ) ∈ [ 0, T ] × S. In view of T is chosen arbitrarily, this completes the proof of uniqueness.
Therefore, (hat{tau}=T), and we complete the proof of the uniqueness part.
