Exercise 13.11: Differentiable Functions and Fourier Coefficients
(a)
By regularity–decay duality of the Fourier transform and the Fourier–Plancherel theorem, \begin{equation} \norm{\F(\partial^{(\alpha)} f)_n}_{\ell^2}^2 \overset{\text{(dual.)}}{=} \norm{(2 \pi i n)^{(\alpha)} \hat f_n}_{\ell^2}^2 \overset{\text{(F.–P.)}}{=} \norm{\partial^{(\alpha)} f}_{L^2[0,1]}^2 \le R. \end{equation} Therefore, \begin{equation} \sum_{n=0}^\infty n^{2 \alpha} (\hat f_n)^2 \lesssim R. \end{equation} Suppose that $(\hat f_n)^2 \gtrsim R n^{-(2 \alpha + 1)}$. Then \begin{equation} R \gtrsim \sum_{n=0}^\infty n^{2 \alpha} (\hat f_n)^2 \gtrsim R \sum_{n=0}^\infty \frac1n, \end{equation} which would mean that the harmonic series is convergent, a contradiction. Therefore, \begin{equation} (\hat f_n)^2 \lesssim R n^{-(2 \alpha + 1)}, \end{equation} which is stronger than required.
(b)
From Example 13.14, we have that \begin{equation} \inf_{f \in \G(1;M)} \norm{f - f^*}_{L^2(\P)}^2 = \sum_{m=M+1}^\infty (\hat f^*_m)^2. \end{equation} Therefore, using (a), \begin{equation} \inf_{f \in \G(1;M)} \norm{f - f^*}_{L^2(\P)}^2 \lesssim R \sum_{m=M+1}^\infty n^{-(2 \alpha + 1)} \le R \int_{M}^\infty x^{-(2 \alpha + 1)} \isd x = \frac{R}{2 \alpha} M^{-2 \alpha}. \end{equation}