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) ).

If you want a strictly positive constant ( C ), take ( f(t) = t ) and look at subsequence ( n = 2k\pi ) not possible, but better: ( f(t)=1 ) fails ( f(0)=0 ). Try ( f(t)=t ): Then ( \limsup n|I_n| = 1 ), so not ( o(1/n) ). If ( f \in C^2 ) and ( f'(0)=0 ) Integrate by parts twice. First as before: [ I_n = \frac1n \int_0^1 f'(t) \cos(nt) dt - \fracf(1)\cos nn. ] Now integrate by parts again on ( J_n := \int_0^1 f'(t) \cos(nt) dt ).

[ 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) ).

Better: By Riemann–Lebesgue lemma, for any ( g \in L^1 ), ( \int g(t) \cos(nt) dt \to 0 ). Here ( g = f' \in L^1 ). Therefore [ \int_0^1 f'(t) \cos(nt) , dt \to 0. ] Hence [ I_n = \frac1n \cdot o(1) = o\left(\frac1n\right). ] Example with ( I_n \sim C/n ) Take ( f(t) = t ). Then ( f(0)=0 ), ( f \in C^1 ). Oraux X Ens Analyse 4 24.djvu

Actually, known result: If ( f ) is ( C^1 ) and ( f(0)=0 ), ( I_n = o(1/n) ). If ( f ) is ( C^2 ) and ( f(0)=f(1)=0 ), then ( I_n = O(1/n^2) ). But here they only give ( f'(0)=0 ), not ( f(1)=0 ). Possibly a misprint? Let's assume they intended ( f(0)=f(1)=0 ) for (3). Then:

I cannot directly access external files such as Oraux X Ens Analyse 4 24.djvu . However, if you provide the text or a specific exercise from that document (e.g., by copying the statement or describing the problem), I can certainly help produce a detailed solution, commentary, or a synthetic correction typical of an oral examination at ENS/X level in analysis.

[ I_n = \left[ -f(t) \frac\cos(nt)n \right]_0^1 + \frac1n \int_0^1 f'(t) \cos(nt) , dt. ] Boundary term: at ( t=1 ): ( -f(1) \frac\cos nn ). At ( t=0 ): ( + f(0) \frac1n = 0 ). So boundary term is ( O(1/n) ). 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.

Let ( u = f'(t) ), ( dv = \cos(nt)dt ), ( du = f''(t) dt ), ( v = \frac\sin(nt)n ).

Compute: [ I_n = \int_0^1 t \sin(nt) dt. ] Integration by parts: ( u = t ), ( dv = \sin(nt)dt ), ( du = dt ), ( v = -\cos(nt)/n ): [ I_n = \left[ -t \frac\cos(nt)n \right]_0^1 + \frac1n \int_0^1 \cos(nt) dt. ] First term: ( -\frac\cos nn ). Second: ( \frac1n \left[ \frac\sin(nt)n \right]_0^1 = \frac\sin nn^2 ). Then [ K_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1

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

Thus [ I_n = -\frac\cos nn + \frac\sin nn^2. ] As ( n \to \infty ), ( I_n = -\frac\cos nn + o\left(\frac1n\right) ). The amplitude of ( I_n ) is ( \sim \frac1n ) up to a bounded oscillatory factor. Indeed ( |I_n| \sim \fracn ), not ( C/n ) with constant sign, but in the sense of equivalence modulo ( o(1/n) ), it's ( O(1/n) ) and not ( o(1/n) ).

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) ).

The integral term: ( \left| \int_0^1 f'(t) \cos(nt) , dt \right| \leq \int_0^1 |f'(t)| dt < \infty ), hence it is bounded. Thus the whole integral term is ( O(1/n) ). Wait — but we need ( o(1/n) ), not just ( O(1/n) ).