Integer Smoothness And Endpoint Conventions

Why Integer Endpoints Need a Convention

For a noninteger smoothness exponent, a Hölder condition typically specifies how the highest derivative changes over a short distance. At an integer exponent, there is no fractional derivative left to constrain in that way. Several natural definitions are then used: requiring continuous derivatives through order $m$, requiring the derivative of order $m-1$ to be Lipschitz, or imposing a finite-difference bound of Zygmund type. These definitions are related, but they are not interchangeable.

This distinction matters when a function class appears in a theorem. A rate statement has both a numerical part—the rate as sample size grows—and a scope: the class of functions over which the risk is bounded or minimized. At an integer endpoint, changing the smoothness convention can change that scope even when the displayed rate exponent is unchanged.

Three Conventions at an Integer Order

Fix an integer $m\geq 1$ and a compact interval $I$. Derivatives at endpoints, when needed, are understood as one-sided derivatives. For $x$ and $h$ such that all arguments lie in $I$, define the forward difference of order $r$ by

\Delta_h^r f(x) = \sum_{j=0}^{r}(-1)^{r-j}\binom{r}{j}f(x+jh).

The most direct convention requires $m$ continuous derivatives. A second convention, common in Hölder classes at integer order, requires the derivative of order $m-1$ to be Lipschitz. A third uses an order-$(m+1)$ difference bounded at order $|h|^m$.

Definition: The class $C^m(I)$ consists of functions with continuous derivatives through order $m$. The class $C^{m-1,1}(I)$ consists of functions with continuous derivatives through order $m-1$ whose $(m-1)$st derivative is Lipschitz. For $m=1$, this means that $f$ itself is Lipschitz. The integer Zygmund class $\Lambda^m(I)$ consists of bounded functions for which $$ \sup_{\substack{h\neq 0,\ x,x+h,\ldots,x+(m+1)h\in I}} \frac{|\Delta_h^{m+1}f(x)|}{|h|^m}<\infty. $$

On a compact interval, $C^m(I)$ is contained in $C^{m-1,1}(I)$: the $m$th derivative is bounded, so the mean value theorem makes the $(m-1)$st derivative Lipschitz. The next result places the latter class inside the Zygmund class. The inclusions are generally strict, as the examples below show.

Worked Example: A Lipschitz Function at Order One

Let $f(x)=x_+=\max(x,0)$ on $[-1,1]$. It is Lipschitz with constant $1$: on either side of zero its slope is $0$ or $1$, and if $x<0<y$, then $|f(y)-f(x)|=y\leq y-x$. It is not differentiable at zero, so it is not in $C^1([-1,1])$.

For $m=1$, the Zygmund condition bounds $|f(x+2h)-2f(x+h)+f(x)|$ by a constant times $|h|$. The inclusion proved next shows that this condition holds for $f$, even though $f\notin C^1$. Thus the continuous-first-derivative convention and the Lipschitz convention already differ at the first integer endpoint.

From a Lipschitz Derivative to a Finite-Difference Bound

Theorem: If $f\in C^{m-1,1}(I)$, then $f\in\Lambda^m(I)$. In particular, if the Lipschitz constant of $f^{(m-1)}$ is $L$, then $$ |\Delta_h^{m+1}f(x)|\leq 2L|h|^m $$ whenever the differences in this expression are defined.

Proof. A Lipschitz function on an interval is absolutely continuous, has an almost-everywhere derivative bounded in absolute value by its Lipschitz constant, and is the integral of that derivative. Apply this fact to $f^{(m-1)}$. It has an almost-everywhere derivative $f^{(m)}$ with $|f^{(m)}|\leq L$. Repeated use of the fundamental theorem of calculus gives, for $h>0$,

\Delta_h^m f(x) = \int_{[0,h]^m} f^{(m)}(x+t_1+\cdots+t_m)\,dt_1\cdots dt_m.

The formula is valid for the absolutely continuous derivatives involved; values of the integrand on a null set do not affect the integral. Taking absolute values bounds this order-$m$ difference by $Lh^m$. An order-$(m+1)$ difference is the difference of two order-$m$ differences:

\Delta_h^{m+1}f(x) = \Delta_h^m f(x+h)-\Delta_h^m f(x).

The triangle inequality therefore gives $|\Delta_h^{m+1}f(x)|\leq 2Lh^m$. For $h<0$, the identity $\Delta_{-h}^r f(x)=(-1)^r\Delta_h^r f(x-rh)$ reduces the bound to the positive-step case, provided all points lie in $I$. This proves the result. $\square$

This proof establishes an inclusion, not an equivalence. The next example exhibits the gap: an integer Zygmund function need not have a Lipschitz derivative of order $m-1$.

Worked Example: A Zygmund Function Outside $C^{m-1,1}$

For any integer $m\geq 1$, set $g(0)=0$ and $g(x)=x^m\log|x|$ for $x\neq 0$, restricted to $[-1,1]$. First, $g$ has continuous derivatives through order $m-1$. Differentiating $k$ times away from zero gives a sum whose terms are constant multiples of $x^{m-k}\log|x|$ and $x^{m-k}$. For $0\leq k\leq m-1$, these terms tend to zero as $x\to 0$, consistent with the derivatives at zero.

To check the Zygmund condition, define $q_m(u)=u^m\log|u|$ for $u\neq0$, with $q_m(0)=0$. For $h>0$, write $x=hu$. Then

\Delta_h^{m+1}g(x) = h^m\Delta_1^{m+1}q_m(u).

Indeed, $g(h(u+j))=h^m(u+j)^m(\log h+\log|u+j|)$. The part involving $\log h$ is a polynomial of degree $m$ in $j$, so its $(m+1)$st difference is zero. It remains to bound $\Delta_1^{m+1}q_m(u)$ uniformly over real $u$.

On any bounded interval of $u$, this difference is bounded: it is a finite sum of values of the continuous function $q_m$, which is bounded on compact intervals. For $|u|>2(m+1)$, the integral formula for a finite difference gives

\Delta_1^{m+1}q_m(u) = \int_{[0,1]^{m+1}} q_m^{(m+1)}(u+t_1+\cdots+t_{m+1})\,dt_1\cdots dt_{m+1}.

Direct differentiation gives $q_m^{(m+1)}(v)=m!/v$ for $v\neq0$. Throughout the integral, $|v|\geq |u|-(m+1)>|u|/2$, so the absolute value of the integral is at most $2m!/|u|$. Thus $\Delta_1^{m+1}q_m$ is bounded for all $u$, and $|\Delta_h^{m+1}g(x)|\leq C_m h^m$ for a finite constant $C_m$. Negative $h$ follows from the difference identity used in the theorem. Hence $g\in\Lambda^m([-1,1])$.

However, for $x\neq0$, the $(m-1)$st derivative has the form $g^{(m-1)}(x)=m!x\log|x|+c_mx$ for a constant $c_m$, and $g^{(m-1)}(0)=0$. Consequently,

\frac{|g^{(m-1)}(x)-g^{(m-1)}(0)|}{|x|} = |m!\log|x|+c_m| \longrightarrow\infty \qquad\text{as }x\to0.

So $g^{(m-1)}$ is not Lipschitz and $g\notin C^{m-1,1}([-1,1])$. This verifies strictness of the inclusion at every integer order, not just at $m=1$.

Another Distinction: Continuous Highest Derivatives

Worked Example: Lipschitz Smoothness Without $m$ Continuous Derivatives

For any integer $m\geq1$, let $f(x)=x_+^m$ on $[-1,1]$. Its derivatives through order $m-1$ are continuous. In particular,

f^{(m-1)}(x)= \begin{cases} m!x,&x>0,\\ 0,&x\leq0. \end{cases}

This derivative is Lipschitz with constant $m!$: on each side of zero the bound follows from its slope, and for $x\leq0<y$, $|f^{(m-1)}(y)-f^{(m-1)}(x)|=m!y\leq m!(y-x)$. But its derivative at zero has left-hand value $0$ and right-hand value $m!$. Thus $f\notin C^m([-1,1])$, while $f\in C^{m-1,1}([-1,1])$ and, by the theorem, $f\in\Lambda^m([-1,1])$.

Together, the examples show that on a compact interval the usual relationships include $C^m\subseteq C^{m-1,1}\subseteq\Lambda^m$, with both inclusions strict in general. Here $C^m\subseteq C^{m-1,1}$ follows from boundedness of the continuous $m$th derivative and the mean value theorem; the second inclusion was proved above. The examples establish that neither reverse inclusion holds in general.

What the Convention Does to a Rate Statement

A minimax rate is always a statement about a specified function class and risk, under specified assumptions on the observations. For instance, in common nonparametric regression settings with integrated squared-error risk, a Hölder-type smoothness parameter $s$ leads to the familiar exponent $n^{-2s/(2s+d)}$, where $d$ is the dimension of the covariate. At an integer value $s=m$, the formula alone does not say whether the class is $C^m$, $C^{m-1,1}$, or $\Lambda^m$.

If upper and lower bounds have been established for a stated endpoint class, one should report that class rather than silently replacing it with another. In particular, the strict inclusions above mean a theorem over $C^{m-1,1}$ is not automatically a theorem over the larger class $\Lambda^m$. Conversely, an upper bound for a larger class also applies to its subclasses, but that observation alone does not establish a matching lower bound for each subclass. The exponent can be the same for several conventions when the approximation and lower-bound arguments support it; sameness of exponents does not make the classes equal.

For example, substituting $s=m$ and $d=1$ into the displayed integrated-risk rate gives $n^{-2m/(2m+1)}$. This is an arithmetic consequence of the rate formula, not a claim that the same theorem applies to every integer smoothness convention. The class, risk, and assumptions must still be checked. For supremum-norm risk, logarithmic factors may also appear, so a statement about the power of $n$ is not necessarily a complete rate description.

Common Mistake

Wrong reasoning: “At integer $m$, Hölder smoothness just means $m$ continuous derivatives, so any result stated for an integer Hölder class applies to $C^m$.” This silently chooses one endpoint convention and may change the function class.
Correction: Read or state the definition. If the condition is a Lipschitz $(m-1)$st derivative, it describes $C^{m-1,1}$; if it bounds $(m+1)$st differences by a constant times $|h|^m$, it describes the integer Zygmund condition. These classes are related by inclusions, not equalities.

Key takeaway: At integer smoothness, specify the endpoint convention alongside the exponent. Derivative and finite-difference definitions can produce different function classes even when the rate exponent under discussion is the same.

Check Your Understanding

Use the definitions and class relations above to answer the following.

For $m=1$, what condition defines $C^{m-1,1}$, and what finite difference defines $\Lambda^m$?
Why does a bounded $m$th derivative imply that the $(m-1)$st derivative is Lipschitz on a compact interval?
In the proof that $x^m\log|x|$ belongs to $\Lambda^m$, why does the term containing $\log h$ disappear?
Why is the $(m-1)$st derivative of $x^m\log|x|$ not Lipschitz at zero?
If a rate theorem is stated for $C^{m-1,1}$, what additional care is needed before claiming it for $\Lambda^m$?

Integer Smoothness And Endpoint Conventions

What You'll Learn

Why Integer Endpoints Need a Convention

Three Conventions at an Integer Order

Worked Example: A Lipschitz Function at Order One

From a Lipschitz Derivative to a Finite-Difference Bound

Worked Example: A Zygmund Function Outside \(C^{m-1,1}\)

Another Distinction: Continuous Highest Derivatives

Worked Example: Lipschitz Smoothness Without \(m\) Continuous Derivatives

What the Convention Does to a Rate Statement

Common Mistake

Check Your Understanding

Integer Smoothness And Endpoint Conventions

What You'll Learn

Why Integer Endpoints Need a Convention

Three Conventions at an Integer Order

Worked Example: A Lipschitz Function at Order One

From a Lipschitz Derivative to a Finite-Difference Bound

Worked Example: A Zygmund Function Outside \(C^{m-1,1}\)

Another Distinction: Continuous Highest Derivatives

Worked Example: Lipschitz Smoothness Without \(m\) Continuous Derivatives

What the Convention Does to a Rate Statement

Common Mistake

Check Your Understanding

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