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Problem 22. Let I : C ([0, 1]; R) Ñ R be defined by
ż1
I(f ) =
f (x)2 dx .
0
Show that I is differentiable at every “point” f P C ([0, 1]; R).
Hint: Figure out what (DI)(f ) is by computing I(f + h) ´ I(f ), where h P C ([0, 1]; R) is
a “small” continuous function.
Remark. A map from a space of functions such as C ([0, 1]; R) to a scalar field such as R
or C is usually called a functional. The derivative of a functional I is usually denoted by
δI instead of DI.
Proof. For each f P C ([0, 1]; R), define Lf (h) = 2
ż1
f (x)h(x)dx.
0
claim: Lf P B(C ([0, 1]; R), R).
Proof of claim: It is trivial that Lf P L (C ([0, 1]; R), R). Let h P C ([0, 1]; R). Then
ˇ
ˇ
ˇLf (h)ˇ ď 2
ż1
ˇ
ˇˇ
ˇ
ˇf (x)ˇˇh(x)ˇdx ď 2}f }8 }h}8 ;
0
thus
ˇ
ˇ
}Lf }B(C ([0,1];R),R) = sup ˇLf (h)ˇ ď 2}f }8 ă 8 .
}h}8 =1
ˇ
ˇ
ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ
= 0.
Claim: lim
}h}8 Ñ0
}h}8
Proof of claim: Since
ż
] ˇ
ˇ
ˇ ˇ 1 [(
)2
ˇ
2
ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ = ˇˇ
f (x) + h(x) ´ f (x) ´ 2f (x)h(x) dxˇ
0
ˇż 1
ˇ
ˇ
ˇ
= ˇ h(x)2 dxˇ ď }h}28 ,
0
by the sandwich lemma we conclude that
ˇ
ˇ
ˇI(f + h) ´ I(f ) ´ Lf (h)ˇ
}h}28
ď lim
= 0.
0 ď lim
}h}8 Ñ0 }h}8
}h}8 Ñ0
}h}8
Therefore, I is differentiable at f , and (DI)(f )(h) = Lf (h).
˝
Problem 30. Let f : R Ñ R be differentiable. Assume that for all x P R, 0 ď f 1 (x) ď f (x).
Show that g(x) = e´x f (x) is decreasing. If f vanishes at some point, conclude that f is
zero.
Proof. To see that g is decreasing, we compute the derivative of g and find that
g 1 (x) = ´e´x f (x) + e´x f 1 (x) = e´x (f 1 (x) ´ f (x)) ď 0 ;
thus g is a decreasing function. Now suppose that f (c) = 0 for some c P R.
1. Since g is decreasing, g(x) ď g(c) = 0 for all x ě c; thus f (x) = ex g(x) = 0 for all
x ě c.
2. Since f 1 (x) ě 0, f is an increasing function, thus f (x) ď f (c) = 0 for all x ď c. Since
f is assumed to be non-negative, we must have f (x) = 0 for all x ď c.
Combining 1 and 2, we conclude that f (x) = 0 for all x P R.
˝
Problem 32. 1. If f : A Ď Rn Ñ Rm and g : B Ď Rm Ñ Rℓ are twice differentiable and
f (A) Ď B, then for x0 P A, u, v P Rn , show that
D2 (g ˝ f )(x0 )(u, v)
(
)
(
)
= (D2 g)(f (x0 )) (Df )(x0 )(u), Df (x0 )(v) + (Dg)(f (x0 )) (D2 f )(x0 )(u, v) .
2. If p : Rn Ñ Rm is a linear map plus some constant; that is, p(x) = Lx + c for some
L P B(Rn , Rm ), and f : A Ď Rm Ñ Rs is k-times differentiable, prove that
(
)(
)
Dk (f ˝ p)(x0 )(u(1) , ¨ ¨ ¨ , u(k) ) = (Dk f ) p(x0 ) (Dp)(x0 )(u(1) ), ¨ ¨ ¨ , (Dp)(x0 )(u(k) .
Proof. 1. First of all, we show that g ˝ f is twice differentiable. Since g and f are both
differentiable, the chain rule implies that g ˝ f is differentiable and
((
)
)
D(g ˝ f )(x) = (Dg)(f (x))(Df )(x) = (Dg) ˝ f (Df ) (x) .
Since g and f are twice differentiable, Dg and Df are differentiable. By the chain rule
(
)
again, (Dg) ˝ f is differentiable; thus the product rule implies that (Dg) ˝ f (Df ) is
differentiable. Therefore, g ˝ f is twice differentiable.
Now by Proposition 6.69 in 共筆, we have
D2 (g ˝ f )(x0 )(u, v) =
n
ÿ
B 2 (g ˝ f )
(x0 )ui vj .
Bx
Bx
j
i
i,j=1
By the chain rule,
m [
ÿ
B(g ˝ f )
Bg
B 2 (g ˝ f )
B ˇˇ
B ˇˇ
Bfk ]
(x0 ) =
(x) =
(f (x))
(x)
ˇ
ˇ
Bxj Bxi
Bxj x=x0 Bxi
Bxj x=x0 k=1 Byk
Bxi
m
[ Bg
ÿ
B ˇˇ
Bfk ]
=
(f (x))
(x)
ˇ
Bxj x=x0 Byk
Bxi
k=1
=
m ÿ
m
m
ÿ
ÿ
B2g
Bfℓ
Bfk
Bg
B 2 fk
(f (x0 ))
(x0 )
(x0 ) +
(f (x0 ))
(x0 ) ;
By
By
Bx
Bx
By
Bx
Bx
ℓ
k
j
i
k
j
i
k=1 ℓ=1
k=1
thus
D2 (g ˝ f )(x0 )(u, v)
n [ ÿ
m
m
]
ÿ
ÿ
B2g
Bg
Bfℓ
Bfk
B 2 fk
=
(f (x0 ))
(x0 )
(x0 ) +
(f (x0 ))
(x0 ) ui vj
Byℓ Byk
Bxj
Bxi
Byk
Bxj Bxi
i,j=1 k,ℓ=1
k=1
n
n
m
(ÿ
)( ÿ
)
ÿ
Bfℓ
Bfk
B2g
(f (x0 ))
(x0 )vj
(x0 )ui
=
Byℓ Byk
Bxj
Bxi
j=1
i=1
k,ℓ=1
m
n
( ÿ
)
ÿ
Bg
B 2 fk
+
(f (x0 ))
(x0 )ui vj .
Byk
Bxj Bxi
i,j=1
k=1
(
)
Letting (Df )(x0 )(w) r denote the r-th component of (Df )(x0 )(w), we obtain that
D2 (g ˝ f )(x0 )(u, v)
m
ÿ
(
)(
)
B2g
=
(f (x0 )) (Df )(x0 )(v) ℓ (Df )(x0 )(u) k
Byℓ Byk
k,ℓ=1
m
ÿ
(
)
Bg
(f (x0 )) D2 f )(x0 )(u, v) k
Byk
k=1
(
)
(
)
= (D2 g)(f (x0 )) (Df )(x0 )u, (Df )(x0 )v + (Dg)(f (x0 )) (D2 f )(x0 )(u, v) .
+
2. The validity of the desired equality for the case k = 1 is the chain rule. Suppose that
the desired holds for k = K. Then for k = K + 1, by Corollary 6.70 in 共筆 we obtain
that
DK+1 (f ˝ p)(x0 )(u(1) , ¨ ¨ ¨ , u(K+1) ) =
n
ÿ
j=1
(K+1)
uj
B ˇˇ
(DK (f ˝ p))(x)(u(1) , ¨ ¨ ¨ , u(K) )
ˇ
Bxj x=x0
ˇ
(
)(
)
(K+1) B ˇ
=
uj
(Dk f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(k) .
ˇ
Bxj x=x0
j=1
n
ÿ
Noting that (Dp)(x)(ur ) = Lu(r) (which is independent of x), by Proposition 6.69 in
共筆 we find that
(
)(
)
(DK f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(K)
n
ÿ
BK f
(p(x))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK ,
=
ByjK ¨ ¨ ¨ Byj1
j ,¨¨¨ ,j =1
1
K
where (Lur )s denotes the s-th component of the vector Lu(r) . As a consequence,
(
)(
)
B ˇˇ
(DK f ) p(x) (Dp)(x)(u(1) ), ¨ ¨ ¨ , (Dp)(x)(u(K)
ˇ
Bxj x=x0
j=1
n
m
ˇ
ÿ
ÿ
BK f
(K+1) B ˇ
=
uj
(p(x))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK
ˇ
Bxj x=x0 j ,¨¨¨ ,j =1 ByjK ¨ ¨ ¨ Byj1
j=1
n
ÿ
(K+1)
uj
1
=
n
ÿ
(K+1)
uj
j=1
=
m
ÿ
j1 ,¨¨¨ ,jK ,jK+1
m
ÿ
j1 ,¨¨¨ ,jK ,jK+1
=
m
ÿ
j1 ,¨¨¨ ,jK ,jK+1
K
Bpj
B K+1 f
(p(x0 )) K+1 (x0 )(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK
ByjK+1 ¨ ¨ ¨ Byj1
Bxj
=1
)
(ÿ
B K+1 f
(K+1) BpjK+1
uj
(x0 ) (Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK
(p(x0 ))
ByjK+1 ¨ ¨ ¨ Byj1
Bxj
j=1
=1
n
(
)
B K+1 f
(p(x0 )) (Dp)(x0 )u(K+1) j (Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK
K+1
ByjK+1 ¨ ¨ ¨ Byj1
=1
m
ÿ
B K+1 f
(p(x0 ))(Lu(1) )j1 ¨ ¨ ¨ (Lu(K) )jK (Lu(K+1) )jK+1
By
¨
¨
¨
By
jK+1
j1
j1 ,¨¨¨ ,jK ,jK+1 =1
(
)(
)
= (DK+1 f ) p(x0 ) (Dp)(x0 )(u(1) ), ¨ ¨ ¨ , (Dp)(x0 )(u(K+1)
=
which shows the validity of the desired equality for the case k = K + 1.
˝
Problem 34. Let f : Rn Ñ Rm be differentiable, and Df is a constant map in B(Rn , Rm );
that is, (Df )(x1 )(u) = (Df )(x2 )(u) for all x1 , x2 P Rn and u P Rn . Show that f is a linear
term plus a constant and that the linear part of f is the constant value of Df .
Proof. Since Df is a constant map, Df is continuous; thus f P C 1 . Therefore, the Taylor
Theorem implies that
f (x) = f (0) + (Df )(c)(x ´ 0)
for some c on the line segment joining x and 0. Let L = (Df )(x). Then
f (x) = f (0) + L(x ´ 0) = Lx + f (0) .
˝
Problem 38. Prove Corollary 7.5; that is, show that if U Ď Rn is open, f : U Ñ Rn is
of class C 1 , and (Df )(x) is invertible for all x P U, then f (W) is open for every open set
W Ď U.
Proof. Let W Ď U be an open set. For each x P W, (Df )(x) is invertible; thus the inverse
function theorem implies that there exists δx ą 0 such that
(
)
(
)
(a) D(x, δx ) Ď W; (b) f D(x, δx ) is open; (c) f : D(x, δ) Ñ f D(x, δx ) is one-to-one
and onto.
Ť
Since W =
D(x, δx ),
xPW
ď (
)
f (W) =
f D(x, δx )
xPW
is the union of infinitely many open sets; thus f (W) is open.
˝
Problem 40. Let f : R2 Ñ R be of class C 1 , and for some (a, b) P R2 , f (a, b) = 0 and
fy (a, b) ‰ 0. Show that there exist open neighborhoods U of a and V of b such that every
y P V corresponds to a unique x P U such that f (x, y) = 0. In other words, there exists a
function y = y(x) such that y(a) = b and f (x, y(x)) = 0 for all x P U.
Proof. Let z = (x, y) and w = (u, v), where x, y, u, v P R. Define w = F (z), where F is
(
)
given by F (x, y) = x, f (x, y) . Then F : D Ñ R2 , and
[
]
1
0
[
]
(DF )(z) =
.
fx (x, y) fy (x, y)
We note that the Jacobian of F at (a, b) is fy (a, b) ‰ 0, so the inverse function theorem
(
)
implies that there exists open neighborhoods O Ď R2 of (a, b) and W Ď R2 of a, f (a, b) =
(a, 0) such that
(a) F : O Ñ W is one-to-one and onto;
(b) the inverse function F ´1 : W Ñ O is of class C r ;
(
) (
)´1
(c) (DF ´1 ) x, f (x, y) = (DF )(x, y) .
W.L.O.G. we can assume that O = U ˆ V, where U Ď R and V Ď R are open, and a P U,
b P V.
(
)
Write F ´1 (u, v) = φ(u, v), ψ(u, v) , where φ : W Ñ U and ψ : W Ñ V. Then
(
) (
)
(u, v) = F φ(u, v), ψ(u, v) = φ(u, v), f (u, ψ(u, v))
(
)
which implies that φ(u, v) = u and v = f (u, ψ(u, v)). Let y(x) = ψ(x, 0). Then u, f (u) P
(
)
U ˆ V is the unique point satisfying f u, y(u) = 0 if u P U. Therefore, f : U Ñ V, and
(
)
f x, y(x) = 0
@x P U .
(
)
(
)
Since G(a, b) = (a, 0) = G a, f (a) , (a, b), a, f (a) P O, and G : O Ñ W is one-to-one, we
must have b = f (a).
˝
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