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  "body": "\n## Statement\n\nLet $f \\colon \\R \\to \\C$ be continuous and $2\\pi$-periodic, let $r \\ge 0$ be an integer, and let $0 \u003c \\alpha \u003c 1$.\nFor $n \\ge 1$, let $E_n(f)$ be the uniform distance from $f$ to the [[trigonometric polynomials|J3NJQ5FZJ5]] of degree at most $n$.\nAssume that there is a constant $M$ with $E_n(f) \\le M n^{-(r + \\alpha)}$ for all $n \\ge 1$.\nThen $f$ has $r$ continuous derivatives, and $f^{(r)}$ is [[Hölder continuous|012PBA0FMD]] with exponent $\\alpha$.\n\n## Remarks\n\nProof idea: let $T_k$ be a best approximation of degree $2^k$ and write $f = T_0 + \\sum_{k \\ge 0} (T_{k+1} - T_k)$.\nEach block $T_{k+1} - T_k$ has degree at most $2^{k+1}$ and size $O(2^{-k(r + \\alpha)})$.\n[[Bernstein's inequality|F0YGC89S9R]] bounds its $j$-th derivative by $O(2^{k(j - r - \\alpha)})$, so the series of $r$-th derivatives converges uniformly.\nSplitting the sum at $2^k \\approx 1/\\abs{x - y}$ gives the Hölder bound for $f^{(r)}$.\n\nExample: if $E_n(f) \\le M n^{-1/2}$ for all $n$, then $\\abs{f(x) - f(y)} \\le M' \\abs{x - y}^{1/2}$.\n\nThe hypothesis $\\alpha \u003c 1$ is necessary.\nFor $\\alpha = 1$ and $r = 0$, the condition $E_n(f) = O(1/n)$ does not imply that $f$ is Lipschitz.\nIt characterises the larger Zygmund class, defined by $\\abs{f(x + h) - 2f(x) + f(x - h)} = O(h)$.\n\n## Explanations\n\n### L1 human\n\nIf a function can be approximated very well by trigonometric polynomials of low degree, it must be smooth.\nThe faster the error goes down, the more derivatives the function has.\nThis is the reverse of Jackson's inequality.\n\n### L2 human\n\nWith Jackson's inequality it gives a two-way bridge: for $0 \u003c \\alpha \u003c 1$, $E_n(f) = O(n^{-r-\\alpha})$ exactly when $f^{(r)}$ is Hölder continuous with exponent $\\alpha$.\nSo smoothness can be read off from approximation rates, which is the starting point of approximation spaces and Besov spaces.\nThe same telescoping argument over dyadic degrees appears in wavelet and Littlewood–Paley theory.\n",
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