Lemma 13 let ? be irrational we can find increasing

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Lemma 13. Let λ be irrational We can find increasing continuous functions φ j : [0 , 1] R [1 j 5] with the following property. Given any continuous function F : [0 , 1] 2 R we can find a function G : R R such that k G k ≤ k F k and sup ( x,y ) [0 , 1] 2 fl fl fl fl fl F ( x, y ) - 5 X j =1 G ( φ j ( x ) + λφ j ( y )) fl fl fl fl fl 999 1000 k F k . Next we make the following observation. Lemma 14. We can find a sequence of functions f n : [0 , 1] 2 R which are uniformly dense in C ([0 , 1]) 2 . This enables us to obtain Lemma 13 from a much more specific result. Lemma 15. Let λ be irrational and let the f n be as in Lemma 14. We can find increasing continuous functions φ j : [0 , 1] R [1 j 5] with the following property. We can find functions g n : R R such that k g n k k f n k and sup ( x,y ) [0 , 1] 2 fl fl fl fl fl f n ( x, y ) - 5 X j =1 g n ( φ j ( x ) + λφ j ( y )) fl fl fl fl fl 998 1000 k f n k . 6
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Now that we have reduced the matter to satisfying a countable set of conditions, we can use a Baire category argument. We need to use the correct metric space. Lemma 16. The space Y of continuous functions φ : [0 , 1] R 5 with norm k φ k = sup t [0 , 1] k φ ( t ) k is complete. The subset X of Y consisting of those φ such that each φ j is increasing is a closed subset of Y . Thus if d is the metric on X obtained by restricting the metric on Y derived from k k we have ( X, d ) complete. Exercise 17. Prove Lemma 16 Lemma 18. Let f : [0 , 1] 2 R be continuous and let λ be irrational. Con- sider the set E of φ X such that there exists a g : R R such that k g k ≤ k f k sup ( x,y ) [0 , 1] 2 fl fl fl fl fl f ( x, y ) - 5 X j =1 g ( φ j ( x ) + λφ j ( y )) fl fl fl fl fl < 998 1000 k f k . The X \ E is a closed set with dense complement in ( X, d ) . (Notice that it is important to take ‘ < ’ rather than ‘ ’ in the displayed formula of Lemma 18.) Lemma 18 is the heart of the proof and once it is proved we can easily retrace our steps and obtain Theorem 11. By using appropriate notions of information Vitushkin was able to show that we can not replace continuous by continuously differentiable in Theo- rem 12. Thus Theorem 11 is an ‘exotic’ rather than a ‘central’ result. 4 The principle of uniform boundedness We start with a result which is sometimes useful by itself but which, for us, is merely a stepping stone to Theorem 22. Lemma 19 (Principle of uniform boundedness). Suppose that ( X, d ) is a complete metric space and we have a collection F of continuous functions f : X R which are pointwise bounded, that is, given any x X we can find a K ( x ) > 0 such that | f ( x ) | ≤ K ( x ) for all f ∈ F . Then we can find a ball B ( x 0 , δ ) and a K such that | f ( x ) | ≤ K for all f ∈ F and all x B ( x 0 , δ ) . 7
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Exercise 20. (i) Suppose that ( X, d ) is a complete metric space and we have a sequence of continuous functions f n : X R and a function f : X R such that f n converges pointwise that is f n ( x ) f ( x ) for all f ∈ F .
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