---
tag: 5V22013VEQ
kind: definition
name: Modulus of smoothness
msc: [41A25, 26A15, 41A17]
defines: CRPZ1RVNKQ
uses: [AXCJBSQZWF, SASP14MA1Z, 0X3WGX95FE]
sources:
  - url: https://en.wikipedia.org/wiki/Modulus_of_smoothness
---

## Definition

Let $a < b$, let $k \ge 1$ be an integer, and let $1 \le p \le \infty$.
Let $f \in L^p[a, b]$, an [[Lp space|SASP14MA1Z]] element; for $p = \infty$ let $f$ be [[continuous|0X3WGX95FE]] on $[a, b]$.
For $h > 0$, the $k$-th forward difference of $f$ with step $h$ is
$$\Delta_h^k f(x) = \sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} f(x + i h), \qquad a \le x \le b - k h.$$
For $0 < t \le (b - a)/k$, the *modulus of smoothness of order $k$* of $f$ is
$$\omega_k(f, t)_p = \sup_{0 < h \le t} \norm{\Delta_h^k f}_{L^p[a, \, b - k h]}.$$

## Remarks

Example: for $f(x) = x$ on $[0, 1]$ and $p = \infty$, one has $\Delta_h^1 f = h$ and $\Delta_h^2 f = 0$, so $\omega_1(f, t)_\infty = t$ and $\omega_2(f, t)_\infty = 0$.

Non-example: the norm of the difference at a single step is not a modulus of smoothness, because it need not grow with $t$.
For $f(x) = \sin(2 \pi x)$ on $[0, 2]$ and $p = \infty$, one has $\Delta_1^1 f = 0$ but $\norm{\Delta_{1/2}^1 f}_\infty = 2$; the supremum over $0 < h \le t$ is what makes $\omega_k(f, \cdot)_p$ nondecreasing.

For $k = 1$ and $p = \infty$, $\omega_1(f, t)_\infty$ is the [[modulus of continuity|AXCJBSQZWF]] of $f$.
One has $\omega_k(f, t)_p \to 0$ as $t \to 0$, and $\omega_{k+1}(f, t)_p \le 2 \omega_k(f, t)_p$.