Corrigé de l`Examen écrit du 24 août 2016

R
R
fRg:x7→ xf(x)
fRDy(F+f)(y) = i(F+g)(y)
y∈R
R2
f g
f(x) = cos(x)χ[−π/2,π/2](x)g(x) = cos (πx/2)
1−x2.
f
g
Z0
−∞
cos2(πx/2)
(1 −x2)2dx.

fR
[−π/2, π/2] y∈R
F±
yf=ZR
e±ixy cos(x)χ[−π/2,π/2](x)dx
= 2 Zπ/2
0
cos(xy) cos(x)dx
=Zπ/2
0
cos((1 −y)x)dx +Zπ/2
0
cos((1 + y)x)dx.
Zπ/2
0
cos((1 −y)x)dx =
hsin((1−y)x)
1−yiπ/2
0y6= 1
Rπ/2
01dx y = 1
=(sin((1−y)π/2)
1−yy6= 1
π
2y= 1.
Zπ/2
0
cos((1 + y)x)dx =
hsin((1+y)x)
1+yiπ/2
0y6=−1
Rπ/2
01dx y =−1
=(sin((1+y)π/2)
1+yy6=−1
π
2y=−1.
sin(π/2−πy/2) = sin(π/2 + πy/2) = cos(πy/2) cos(π/2) = cos(−π/2) = 0
F±
yf=(2 cos(πy/2)
1−y2y∈R\{−1,1}
π
2y∈ {−1,1}.
gR\{−1,1}g=1
2F±fR\{−1,1}
RgR
limx→±∞ |x|3/2|g(x)|= 0 gR
g|g(x)| ≤ 2/x2
x∈]− ∞,−√2[ ∪]√2,+∞[x7→ 2/x2−∞ +∞
y∈R
F±
yg=1
2F±
yF∓f=πf(y) = πcos(y)χ[−π/2,π/2](y)
fR
g2]−∞,0[
Rlimx→−∞ x3g2(x) = 0
Z0
−∞
cos2(πx/2)
(1 −x2)2dx =1
2ZR
g2(x)dx
=1
8ZRF+
xfF− xfdx
=1
8ZR
f(x)F+
xF−fdx
=π
4ZR
f2(x)dx
=π
2Zπ/2
0
cos2(x)dx
=π
2Zπ/2
0
1 + cos(2x)
2dx
=π
4x+sin(2x) 2π/2
0
=π2
8.

f g
f(x, y) = ln (ln(x)) −ln(xy)+2y2g(x, y) = x2+ 4y2.
f
f g
g
g
A=n(x, y)∈R2:y≥0px2+y2≤1o.
f C2
B:= ]1,+∞[×]0,+∞[,
f
Dxf(x, y)=0
Dyf(x, y)=0 ⇔
1
xln(x)−1 x= 0
−1
y+ 4y= 0 ⇔(ln(x)=1
y2=1
4⇔(x=e
y=±1
2.
f
(e, 1/2) f
D2
xf(x, y) = −ln(x)−1
x2ln2(x)+1
x2, DxDyf(x, y)=0 D2
yf(x, y) = 1
y2+ 4,
Hf(e, 1/2) = −1/e20
0 8 det (Hf(e, 1/2)) = −8/e2<0.
(e, 1/2) f f
g g ∈C2(R2)Dxg(x, y)=2x Dyg(x, y)=8y
g(0,0) g
g C (x, y)∈C
x2+y2= 1 g(x, y) = 3y2+1
g0:y7→ 3y2+ 1 [−1,1] g0
y= 0 y=−1y= 1 g(1) = g(−1) = 4
C g (0,−1) (0,1)
(−1,0) (1,0)
A

g
A
A g
g(0,1)
(−1,0) (1,0)
(x, 0) x∈[−1,1] g(x, 0) = x2/2
g(0,0) (−1,0) (1,0)
(0,0) gR2
A(1,0) (−1,0)
(0,1) g
A