# Uniform Lipshitz continuity implies Continuous Differentiability

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 $α-1$.
• 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 $h = v-u$? – abhi01nat Nov 25 '17 at 7:16
• 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 $C_1$. – LutzL Nov 25 '17 at 8:05
• 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. – LutzL Nov 25 '17 at 9:38
• 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 – LutzL Nov 25 '17 at 22:27