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  "body": "\n## Statement\n\nLet $n \\ge 2$ and $1 \\le k \u003c n$ be integers.\nThere is a constant $C(n, k)$, depending only on $n$ and $k$, with the following property.\nLet $f \\colon \\R \\to \\R$ be $n$ times differentiable, with $f$ and its $n$-th [[derivative|084NGSTWM7]] $f^{(n)}$ bounded on $\\R$.\nThen $f^{(k)}$ is bounded on $\\R$ and\n$$\\norm{f^{(k)}}_\\infty \\le C(n, k) \\, \\norm{f}_\\infty^{1 - k/n} \\, \\norm{f^{(n)}}_\\infty^{k/n},$$\nwhere $\\norm{g}_\\infty = \\sup_{x \\in \\R} \\abs{g(x)}$ is the [[supremum|0BTMQ04556]] of $\\abs{g}$.\n\n## Remarks\n\nProof idea for $n = 2$ and $k = 1$: let $M_0 = \\norm{f}_\\infty$ and $M_2 = \\norm{f''}_\\infty$.\nBy [[Taylor's theorem|03QHBVW4TT]], $f(x \\pm h) = f(x) \\pm h f'(x) + \\frac{h^2}{2} f''(\\xi_\\pm)$ for some $\\xi_\\pm$.\nSubtracting gives $\\abs{f'(x)} \\le \\frac{M_0}{h} + \\frac{h M_2}{2}$ for every $h \u003e 0$, and the choice $h = \\sqrt{2 M_0 / M_2}$ gives $\\norm{f'}_\\infty^2 \\le 2 M_0 M_2$.\nThe constant $2$ here is the best possible on $\\R$.\nOn the half-line $[0, \\infty)$ the same kind of bound holds with the best constant $4$ in place of $2$.\n\nThe exponents are forced by scaling: replacing $f(x)$ by $f(\\lambda x)$ multiplies $\\norm{f^{(j)}}_\\infty$ by $\\lambda^j$.\n\n## Explanations\n\n### L1 human\n\nIf a function stays small and its $n$-th derivative also stays small, then every derivative in between must stay small too.\nA function cannot wiggle fast without either growing large or having a large high-order derivative.\nThe inequality gives the precise trade-off.\n\n### L2 human\n\nThese are interpolation inequalities: the middle derivative is controlled by a weighted geometric mean of the two ends.\nThey are a model for the Gagliardo–Nirenberg inequalities for Sobolev norms.\nThe best constants $C(n, k)$ on $\\R$ are known for all $n$ and $k$, and they are attained by functions related to Euler splines.\nFor $n = 2$ and $k = 1$ the best constant is $\\sqrt{2}$.\n",
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