Oraux X Ens Analyse 4 24.djvu Apr 2026

Integrate by parts twice: First: ( I_n = \frac1n \int_0^1 f'(t)\cos(nt) dt ) (boundary term vanishes because ( f(0)=f(1)=0 )). Second: Let ( K_n = \int_0^1 f'(t)\cos(nt) dt ). Integrate by parts: ( u = f'(t) ), ( dv = \cos(nt) dt ), ( du = f''(t) dt ), ( v = \sin(nt)/n ). Then [ K_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1 - \frac1n \int_0^1 f''(t) \sin(nt) dt. ] Boundary term: at ( t=1 ), ( f'(1)\sin n /n = O(1/n) ); at ( t=0 ), ( f'(0)\sin 0 / n = 0 ). So ( K_n = O(1/n) ). Then [ I_n = \frac1n \cdot O\left(\frac1n\right) = O\left(\frac1n^2\right). ] With ( f'' ) integrable, the remaining integral ( \int f''(t)\sin(nt) dt \to 0 ) by Riemann–Lebesgue, giving ( o(1/n^2) ).

[ J_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1 - \frac1n \int_0^1 f''(t) \sin(nt) dt. ] Boundary: at ( t=1 ): ( f'(1) \sin n / n ); at ( t=0 ): ( f'(0) \cdot 0 / n = 0 ). So ( J_n = O(1/n) ). Oraux X Ens Analyse 4 24.djvu

We made a mistake: The boundary term at ( t=0 ) in the second integration by parts: ( f'(0) \sin(0)/n = 0 ) indeed, but the first integration by parts gave the term ( -f(1)\cos n / n ). That term is ( O(1/n) ), not smaller. So we cannot get ( o(1/n^2) ) unless ( f(1)=0 ). But the problem didn't assume ( f(1)=0 ). Possibly the intended condition is ( f(0)=f(1)=0 ) and ( f'(0)=0 )? Or perhaps the statement in (3) is: prove ( I_n = o(1/n) ) (already done) but with ( C^2 ) and ( f'(0)=0 ) we can improve? Wait, let's recompute properly with a view to ( o(1/n^2) ). Integrate by parts twice: First: ( I_n =

Thus [ I_n = \frac1n J_n - \fracf(1)\cos nn = \frac1n \left( O(1/n) \right) - \fracf(1)\cos nn = -\fracf(1)\cos nn + O\left(\frac1n^2\right). ] So ( I_n = O(1/n) ), not yet ( o(1/n^2) ). Hmm — but the problem statement says: if ( f'(0)=0 ) and ( f \in C^2 ), prove ( I_n = o(1/n^2) ). That suggests extra cancellation in the boundary term? Let's check carefully. Then [ K_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1

Thus ( I_n = o(1/n^2) ).