Bernstein's theorem on inverse approximation

Remarks

Proof idea: let $T_k$ be a best approximation of degree $2^k$ and write $f = T_0 + \sum_{k \ge 0} (T_{k+1} - T_k)$. Each block $T_{k+1} - T_k$ has degree at most $2^{k+1}$ and size $O(2^{-k(r + \alpha)})$. Bernstein's inequality bounds its $j$-th derivative by $O(2^{k(j - r - \alpha)})$, so the series of $r$-th derivatives converges uniformly. Splitting the sum at $2^k \approx 1/\abs{x - y}$ gives the Hölder bound for $f^{(r)}$.

Example: if $E_n(f) \le M n^{-1/2}$ for all $n$, then $\abs{f(x) - f(y)} \le M' \abs{x - y}^{1/2}$.

The hypothesis $\alpha < 1$ is necessary. For $\alpha = 1$ and $r = 0$, the condition $E_n(f) = O(1/n)$ does not imply that $f$ is Lipschitz. It characterises the larger Zygmund class, defined by $\abs{f(x + h) - 2f(x) + f(x - h)} = O(h)$.

Going deeper

For a student

With Jackson's inequality it gives a two-way bridge: for $0 < \alpha < 1$, $E_n(f) = O(n^{-r-\alpha})$ exactly when $f^{(r)}$ is Hölder continuous with exponent $\alpha$. So smoothness can be read off from approximation rates, which is the starting point of approximation spaces and Besov spaces. The same telescoping argument over dyadic degrees appears in wavelet and Littlewood–Paley theory.