Modulus of smoothness

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 of $f$. One has $\omega_k(f, t)_p \to 0$ as $t \to 0$, and $\omega_{k+1}(f, t)_p \le 2 \omega_k(f, t)_p$.