Mallat in his book on wavelets makes the following definition - A function f:[a,b]→R is (C,α)-Lipschitz at v∈[a,b] if there is a polynomial pv of degree at most ⌊α⌋ such that |f(x)−pv(x)|≤C|x−v|α for any x∈[a,b]. It is uniformly Lipschitz if it is Lipschitz for all v∈[a,b] with a constant that is independent of v. Continuous differentiability implies Lipschitz continuity as above, since one can use the Taylor polynomial of the function.
Question: Apparently, the converse is also true - uniform (C,α)-Lipschitz continuity implies that the function is ⌊α⌋ times continuously differentiable. I'm having difficulty proving this. It is clear that if α>1 then the function is differentiable. I'm guessing the derivative is uniformly α−1 Lipschitz which would complete the proof by induction, but I'm having trouble showing this. Any hints would be appreciated!
Answer
There exist difference formulas Lu(p)=1h∑k∈Iwkp(u+kh) for some fixed finite index set I⊂Z (usually I=0,1,...,n) and constants wk that are yield Lu(p)=p′(u) for all polynomials up to degree n. Especially from the formula for p≡1 we get ∑k∈Iwk=0. Then as \begin{align} |p_v(x)-p_u(x)|&\le|p_v(x)-f(x)|+|f(x)-p_u(x)|\le C|x-v|^α+C|x-u|^α \end{align} we get \begin{align} |p_v'(u)-p_u'(u)|&=|L_u(p_v)-L_u(p_u)| \\ &\le\frac1h\sum_{k\in I}|w_k|\,|p_v(u+kh)-p_u(u+kh)| \\ &\le\frac1h\sum_{k\in I}|w_k|\,C(|u+kh-v|^α+|kh|^α) \end{align} Now p_u'(u)=f'(u) and for fixed v p_v'(u)=L_u(p_v)=\frac1h\sum_{k\in I}w_kp_v(u+kh)) =\sum_{k\in I}w_k\frac{p_v(u+kh)-p_v(u)}{h} is a polynomial in u on the left side and thus also on the right, all occurrences of h cancel, including the division. To get the desired bound, set h to a multiple of v-u, to get a reasonably small constant, set nh=v-u so that you get |f'(u)-p_v'(u)|\le C_1|u-v|^{α-1} with C_1=nC\sum_{k\in I}|w_k|\,(|1-\tfrac{k}{n}|^α+|\tfrac{k}n|^α). This establishes the assumptions for f' with Hölder/Lipschitz index α-1.
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