Thursday, August 8, 2019

real analysis - Uniform Lipshitz continuity implies Continuous Differentiability


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|xv|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 α1 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)=1hkIwkp(u+kh) for some fixed finite index set IZ (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 p1 we get kIwk=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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