Mallat in his book on wavelets makes the following definition - A function $f : [a,b] \to \mathbb{R}$ is $(C, \alpha)$-Lipschitz at $v \in [a, b]$ if there is a polynomial $p_v$ of degree at most $\lfloor \alpha \rfloor$ such that $|f(x) - p_v(x)| \leq C|x - v|^\alpha$ for any $x \in [a, b]$. It is uniformly Lipschitz if it is Lipschitz for all $v \in [a ,b]$ 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 $(C, \alpha)$-Lipschitz continuity implies that the function is $\lfloor \alpha \rfloor$ times continuously differentiable. I'm having difficulty proving this. It is clear that if $\alpha > 1$ then the function is differentiable. I'm guessing the derivative is uniformly $\alpha-1$ Lipschitz which would complete the proof by induction, but I'm having trouble showing this. Any hints would be appreciated!


There exist difference formulas $$ L_u(p)=\frac1h\sum_{k\in I}w_kp(u+kh) $$ for some fixed finite index set $I\subset\Bbb Z$ (usually $I=0,1,...,n$) and constants $w_k$ that are yield $$ L_u(p)=p'(u) $$ for all polynomials up to degree $n$. Especially from the formula for $p\equiv1$ we get $$\sum_{k\in I}w_k=0.$$ 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 $α-1$.

  • $\begingroup$ Thanks for your answer! One clarification - even before setting $nh = v-u$, since the left hand side is a polynomial in $u$ isn't the right hand side also independent of $h$? Also, why do we set $nh = v-u$ instead of just $h = v-u$? $\endgroup$ – abhi01nat Nov 25 '17 at 7:16
  • 1
    $\begingroup$ In principle yes, then one does not need $q_v$ as that is $p_v'$. $L_u$ is only needed for the estimates. In principle you could use any multiple of $v-u$, I was aiming for a smallish $C_1$. $\endgroup$ – Lutz Lehmann Nov 25 '17 at 8:05
  • $\begingroup$ Edited to remove $q_v$. Could have used from the start $v+k/n(u-v)$ as points, but this would just repeatedly show non-relevant information. $\endgroup$ – Lutz Lehmann Nov 25 '17 at 9:38
  • $\begingroup$ Also, where could I find these difference formulas? That's a neat property for polynomials, I haven't seen that before. Could you provide an example, say for degree 2? $\endgroup$ – abhi01nat Nov 25 '17 at 20:45
  • $\begingroup$ In general the coefficients are solutions of a linear system involving the vanderMonde matrix. For degree 2 you get the usual backward differentiation formulas $f'(x)=\dfrac{-3f(x)+4f(x+h)-f(x+2h)}{2h}+O(h^2)$, which are exact for quadratic $f$ with linear derivative, see for instance here: math.stackexchange.com/a/2507239/115115 $\endgroup$ – Lutz Lehmann Nov 25 '17 at 22:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.