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  "body": "\n## Definition\n\nLet $a \u003c b$, let $k \\ge 1$ be an integer, and let $1 \\le p \\le \\infty$.\nLet $f \\in L^p[a, b]$, an [[Lp space|SASP14MA1Z]] element; for $p = \\infty$ let $f$ be [[continuous|0X3WGX95FE]] on $[a, b]$.\nFor $h \u003e 0$, the $k$-th forward difference of $f$ with step $h$ is\n$$\\Delta_h^k f(x) = \\sum_{i=0}^{k} (-1)^{k-i} \\binom{k}{i} f(x + i h), \\qquad a \\le x \\le b - k h.$$\nFor $0 \u003c t \\le (b - a)/k$, the *modulus of smoothness of order $k$* of $f$ is\n$$\\omega_k(f, t)_p = \\sup_{0 \u003c h \\le t} \\norm{\\Delta_h^k f}_{L^p[a, \\, b - k h]}.$$\n\n## Remarks\n\nExample: 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 $\\omega_2(f, t)_\\infty = 0$.\n\nNon-example: the norm of the difference at a single step is not a modulus of smoothness, because it need not grow with $t$.\nFor $f(x) = \\sin(2 \\pi x)$ 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 $0 \u003c h \\le t$ is what makes $\\omega_k(f, \\cdot)_p$ nondecreasing.\n\nFor $k = 1$ and $p = \\infty$, $\\omega_1(f, t)_\\infty$ is the [[modulus of continuity|AXCJBSQZWF]] of $f$.\nOne has $\\omega_k(f, t)_p \\to 0$ as $t \\to 0$, and $\\omega_{k+1}(f, t)_p \\le 2 \\omega_k(f, t)_p$.\n",
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