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By Constantin Banica, Octavian Stanasila

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4 (Weissinger). Let C be a (nonempty) closed subset of a Banach space X. 21) < ∞. Then K has a unique fixed point x such that   ∞ K n (x) − x ≤  θj  K(x) − x , x ∈ C. 22) j=n Our first objective is to give some concrete values for the existence time T0 . 3. 5 (Picard-Lindel¨of). Suppose f ∈ C(U, Rn ), where U is an open subset of Rn+1 , and f is locally Lipschitz continuous in the second argument. Choose (t0 , x0 ) ∈ U and δ > 0, T > t0 such that [t0 , T ] × Bδ (x0 ) ⊂ U . 23) t0 x∈Bδ (x0 ) L(t) = |f (t, x) − f (t, y)| .

16. Suppose U = R × Rn and for every T > 0 there are constants M (T ), L(T ) such that |f (t, x)| ≤ M (T ) + L(T )|x|, (t, x) ∈ [−T, T ] × Rn . 11) are defined for all t ∈ R. Proof. Using the above estimate for f we have (t0 = 0 without loss of generality) t |φ(t)| ≤ |x0 | + (M + L|φ(s)|)ds, t ∈ [0, T ] ∩ I. 7) M LT (e − 1). 68) L Thus φ lies in a compact ball and the result follows by the previous lemma. 69) where M (t), L(t) are locally integrable. 15. 16 is false (in general) if the estimate is replaced by |f (t, x)| ≤ M (T ) + L(T )|x|α with α > 1.

All rational numbers in this set). Since fn (x1 ) is bounded, we can choose a subsequence (1) (1) fn (x) such that fn (x1 ) converges (Bolzano-Weierstraß). Similarly we can (2) (1) extract a subsequence fn (x) from fn (x) which converges at x2 (and hence (1) also at x1 since it is a subsequence of fn (x)). By induction we get a se(j) (n) quence fn (x) converging at x1 , . . , xj . ). We will show that it converges uniformly for all x: Fix ε > 0 and choose δ such that |fn (x) − fn (y)| ≤ 3ε for |x − y| < δ.

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