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Assuming that ∫ 0 2 π f ( s ) d s = 0, ∫ 0 T h ( t ) d t = 0, (2). it is not difficult to prove that Ψ achieves its infimum on H T 1 and, consequently, that (1) is solvable.
It is not difficult to prove the following statements.
It is not difficult to prove the following lemma.
It is not difficult to prove the following lemma by using the method of mathematical induction.
Using Theorem 2.1, it is not difficult to prove the following.
Then, similarly to Proposition 1, it is not difficult to prove the following.
So, it is not difficult to prove that is compact in and is continuous.
Note that, for all and above formula, it is not difficult to prove (3.5).
It is not difficult to prove this result by using the maximum norm.
It is not difficult to prove that the conditions (i - iii) are sati - iii
It is not difficult to prove that f satisfies the Hartman-Nagumo condition.
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