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It's a non-homogeneous linear equation with constant coefficients, that you can solve by the variation of constant formula. In complex notation, let $z(t):=f(\frac{t}{n})+ \frac{i}{n}f'(\frac{t}{n})$ for $t\in[0,n]$, and let $h(t):= \frac{1}{n^2}(\sin \frac{\theta t}{n} )^n$. So $z(0)=0$, and $\dot z(t)+iz(t)=ih(t)\ ,$ whence $$z(t)=i \int_ 0 ^ t h(s) e^ {i (s-t)} ds \ ,$$ that one can re-write in terms of $f$ and $f'$. This already shows a row estimate $\|f\|_{\infty,[0,1]}\leq 1/n$ $\|f\|_{\infty,[0,1]}\leq 1/n$$and \|f'\| \|f'\| _ {\infty,[0,1]}\leq 11\ ,$$ just because$\|h\| _ \infty\leq 1/n^2.$As a better estimate,$|f(t)| $|f(t)| \le\frac{1}{n} \int_0^1 | \sin(\theta s)|^n ds = O(n^{-3/2})O(n^{-3/2})$$and$$|f'(t)| \le \int_0^1 | \sin(\theta s)|^n ds = O(n^{-1/2})\ .$ $$2 added 109 characters in body It's a non-homogeneous linear equation with constant coefficients, that you can solve by the variation of constant formula. In complex notation, let z(t):=f(\frac{t}{n})+ \frac{i}{n}f'(\frac{t}{n}) for t\in[0,n], and let h(t):= \frac{1}{n^2}(\sin \frac{\theta t}{n} )^n. So z(0)=0, and \dot z(t)+iz(t)=ih(t)\ ,  whence$$z(t)=\int_ $z(t)=i \int_ 0 ^ t h(s) e^ {i (t-s)} s-t)} ds \ ,$$that one can re-write in terms of f and f'. This already shows a row estimate \|f\|_{\infty,[0,1]}\leq 1/n and \|f'\| _ {\infty,[0,1]}\leq 1, just because \|h\|\leq \|h\| _ \infty\leq 1/n^2. As a better estimate, |f(t)| \le\frac{1}{n} \int_0^1 | \sin(\theta s)|^n ds = O(n^{-3/2})\ . 1 It's a non-homogeneous linear equation with constant coefficients, that you can solve by the variation of constant formula. In complex notation, let z(t):=f(\frac{t}{n})+ \frac{i}{n}f'(\frac{t}{n}) for t\in[0,n], and let h(t):= \frac{1}{n^2}(\sin \frac{\theta t}{n} )^n. So z(0)=0, and \dot z(t)+iz(t)=ih(t)\ , whence$$z(t)=\int_ 0 ^ t h(s) e^ {i (t-s)} ds \ ,$$that one can re-write in terms of$f$and$f'$. This already shows a row estimate$\|f\|_{\infty,[0,1]}\leq 1/n$and$\|f'\| _ {\infty,[0,1]}\leq 1$, just because$\|h\|\leq 1/n^2.\$