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We call (K, (F k )) a GIFS Given ck , the following construction yields an attractor i which is the graph of a continuous function f , that interpolates a set of given points i {( m , yi ), i = 0, , m} (for simplicity, we consider the case m = 2, although the general case can be treated in a similar way) Consider the graph of a non-af ne continuous function on [0; 1], we note: (0) = u, (1) = v
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then we choose ak and bk so that the following conditions hold For i = 0, 1: i i
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1 Si (0, u) =
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i i+1 1 , yi , Si (1, v) = , yi+1 m m
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2 2 2 2 S0 (0, y0 ) = (0, y0 ), S0 (1/2, y1 ) = S1 (0, y0 ), S1 (1/2, y1 ) = (1/2, y1 ) 2 2 2 2 S2 (1/2, y1 ) = (1/2, y1 ), S2 (1, y2 ) = S3 (1/2, y1 ), S3 (1, y2 ) = (1, y2 )
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For k > 2 and i = 0, , 2k 1: 1) if i is even, then: a) if i < 2k 1 :
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] (0, y0 ) ] (0, y0 )
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] (1/2, y1 )
k = Si+1 S[k 1 ] S[k 2 ] S[2 i+1 ] (0, y0 ) i+1 i+1
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b) if i 2
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k Si S k 1 S[k 2 S[2 i i ]
] (1/2, y1 ) ] (1/2, y1 )
= S k 1 S[k 2 S[2 i i ]
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i 2k 2
k Si S k 1 S[k 2 S[2 i i ]
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] (1, y2 )
k = Si+1 S[k 1 ] S[k 2 ] S[2 i+1 ] (1/2, y1 ) i+1 i+1
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2) if i is odd, then: a) if i < 2k 1 :
k Si S[k 1 S[k 2 S[2 i i ] ]
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] (1/2, y1 ) ] (1/2, y1 )
= S[k 1 S[k 2 S[2 i i ] ]
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b) if i 2k 1 :
k Si S[k 1 S[k 2 S[2 i i ] ]
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] (1, y2 ) ] (1, y2 )
= S[k 1 S[k 2 S[2 i i ] ]
2 22
i 2k 2
This set of conditions, which we call continuity conditions, ensures that f is a i continuous function that interpolates the points ( m , yi ) The H lder function f is given by the following proposition PROPOSITION 94 Let us assume that the conditions (c) and (c ) are satis ed Then, the attractor of the GIFS, de ned above, is the graph of a continuous function f such that: i = yi i = 0, , m f m and: f (t) = min( 1 , 2 , 3 )
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where: log ck k 1 i1 +mk 2 i2 ++mik 1 +ik c2 1 +i2 c11 mi i m = lim inf 1 k + log(m k ) log ck k 1 j1 +mk 2 j2 ++mjk 1 +jk c2 1 +j2 c11 mj j m 2 = lim inf k + log(m k ) log ck k 1 l1 +mk 2 l2 ++mlk 1 +lk c2 1 +l2 c11 ml l m 3 = lim inf k + log(m k ) and where ip , jp and lp are de ned as in Proposition 91
(92)
1 NOTE 91 Given m real numbers u1 , , um ] m ; 1[, let us de ne, for any k 1 k k and for any i {0, , m 1}, ci as:
i ck = ri+1 m[ m ] i
In this case, we recover the original construction of the usual IFS, considered in section 93 We now prove that GIFS allow us to solve the problem of characterizing the H lder functions Indeed, we have the following main result THEOREM 91 Let s be a function of [0; 1] in [0; 1] The following conditions are equivalent: 1) s is the H lder function of a continuous function f of [0; 1] in ; 2) there is a sequence (sn )n 1 of continuous functions such that: s(x) = lim inf sn (x),