MATH 404 ANALYSIS II— MID SEMESTER TEST-SOLUTIONS
Problem 1.
(a) Let A = {(x, y) ∈ R2 | 0 y < x2 }. Determine A◦ , Ā, and ∂A.
(b) Give an example of countably many closed subsets of R whose
union is an open subset of R.
Solution: (a) A◦ = {(x, y) ∈ R2 | 0 < y < x2 }, Ā = {(x, y) ∈ R2 | 0 y x2 }, ∂A = {(x, y) ∈ R2 | y = 0 or y = x2 }.
(b)
For example, take Sn = [1/n1 − 1/n] for n ≥ 2. Then S =
n≥2 Sn = (0, 1).
Problem 2.
(a) Let a ∈ Rn and let f : Rn → R be defined by f (x) = |x − a|. Show
that f is continuous on Rn .
(b) Show that f = (f1 , . . . , fm ) : Rn → Rm is continuous at a if and
only if each component function fi : Rn → R is continuous at a.
Solution: (a) The continuity of f follows from
|f (x) − f (y)| = ||x − a| − |y − a|| |(x − a) − (y − a)| = |x − y|.
(b) Since |fi (x) − fi (a)| |f (x) − f (a)|, the continuity of f implies
that each fi is continuos at a. Conversely, if every fi is continuous,
√
then given ε > 0, there is δ > 0 so that |fi (x) − fi (a)| < ε/ n for all
|x − a| < δ. So, if |x − a| < δ, then
2
|f (x) − f (a)| =
n
√
|fi (x) − fi (a)| < n(ε/ n)2 = ε2
i=1
showing that f is continuous.
Problem 3.
(a) Let f : Rn → Rm and a ∈ Rn . State the definition of differentiability of f at a.
(b) A function f : Rn × Rm → Rp is called bilinear if for x, x1 , x2 ∈ Rn ,
y, y1, y2 ∈ Rm and α ∈ R we have
f (αx, y) = αf (x, y) = f (x, αy)
f (x1 + x2 , y) = f (x1 , y) + f (x2 , y)
f (x, y1 + y2 ) = f (x, y1 ) + f (x, y2 ).
1
Show:
|f (h, k)|
= 0.
|(h, k)|
(ii) Df (a, b)(h, k) = f (a, k) + f (h, b).
Solution: (i) Let C = max{|f (ei , ej )||1 i n, 1 j m} where
basis in Rn and
e1 , . . . , em is the standard
e1 , . . . , en is the standard
n
m
basis in Rm . Take h = i= hi ei and k = j=1 kj ej . Since f is bilinear,
we have
|f (h, k)| = |
hi kj f (ei , ej )| C
|hi ||kj | C
|h||k|
(i) lim(h,k)→(0,0)
i,j
C
|h|2 + |k|2
2
i,j
i,j
=
i,j
Cnm
(|h|2 + |k|2 ) = C (|h|2 + |k|2 ).
2
So,
|f (h, k)|
C |h|2 + |k|2 → 0
|(h, k)|
as (h, k) → (0, 0).
(ii) Since f is bilinear,
f (a + h, b + k) = f (a, b) + f (a, k) + f (h, k) + f (h, k)
so that
|f (a + h, b + k) − f (a, b) − [f (a, k) + f (h, k)]|
|f (h, k)|
=
→ 0.
|(h, k)|
|(h, k)|
Since (h, k) → f (a, k)+f (h, b) is linear, it follows that Df (a, b)(h, k) =
f (a, k) + f (h, b).
Problem 4.
(a) Let f : Rn → R. Define a term: the function f is of class C 1 .
(b) Show that the function f : R2 → R defined by f (x, y) = |xy| is
differentiable at (0, 0) but is not of class C 1 in any neighborhood of
(0, 0).
Solution: (b) Observe that f (0, y) = 0 and f (x, 0) = 0. This implies
that Dx f (0, 0) = limt→0 [f (t, 0) − f (0, 0)]/t = 0 = Dy f (0, 0). Hence if
f is differentiable, then Df (0, 0) = [0 0]. Since
x2 + y 2
f (x, y) − f (0, 0) − 0 · x − 0 · y
|xy|
1 x2 + y 2
→0
=
·
=
2 x2 + y 2
2
x2 + y 2
x2 + y 2
as (x, y) → 0. Hence, f is indeed differentiable at (0, 0) and Df (0, 0) =
[0 0]. For (0, y) ∈ Br (0, 0) and x = 0, we have
|x||y|
f (x, y) − f (0, y)
=
= ±|y|
x
x
so that the limit as x → 0 does not exists. This implies that Dx f (0, y)
does exists for every (0, y) with y = 0. In particular, f is not of class
C 1 in any neighborhood of (0, 0).
Problem 5.
(a) State the inverse function theorem.
(b) Let f : U ⊂ Rn → Rn be a C 1 function defined on an open subset U of Rn . Assume that f is one-to-one and that detDf (x) = 0 for
every x ∈ U. Show that V = f (U) is an open subset of Rn and that
f −1 : V → Rn is differentiable.
Solution: (b) It suffices to show that if b ∈ V = f (U), then there
is r > 0 so that Br (b) ⊂ V . There is a unique a ∈ U such that
b = f (a). By assumption, detDf (a) = 0. So, by the inverse function theorem, there an open set A ⊂ U containing a and an open
set B containing b = f (a) such that f : A → B has an inverse
function f −1 : B → A which is of class C 1 . For every y ∈ B,
f −1 (y) ∈ A ⊂ U and f (f −1 (y)) = y so that B ⊂ f (A) ⊂ f (U) = V .
Since b ∈ Br (b) ⊂ B ⊂ U for some r > 0 (since B is open and contain
b), it follows that every point in V is an interior point and, therefore,
V is open. Now, since f is one-to-one, there exists the inverse function
f −1 : V = f (U) → U which is of class C 1 by the inverse function
theorem.
Problem 6.
Let f : R3 → R2 be of class C 1 . Write f
Assume that f (3, −1, 2) = 0 and
1
2
Df (3, −1, 2) =
1 −1
in the form f (x, y1 , y2).
1
.
1
(a) Show that there exists a neighborhood U ⊂ R of 3 and a C 1 function g : U → R2 so that g(3) = (−1, 2) and f (x, g1 (x), g2 (x)) = 0
for all x ∈ U.
(b) Find Dg(3).
(c) Can one solve the equation f (x, y1 , y2 ) = 0 for (x, y1 ) in terms of
y2 ? If yes, why?
2 1
= 3. So,
Solution: (a) Calculate detDy f (3, −1, 2) = det
−1 1
by the implicit function theorem there exists a neighborhood U of 3
and a C 1 -function g = (g1 , g2) : U → R2 such that g(3) = (−1, 2) and
f (x, g1 (x), g2 (x)) = 0 for all x ∈ U.
(b) The derivative Dg(3) is equal to
Dg(3) = −[Dy f (3, −1, 2)]−1Dx f (3, −1, 2).
The inverse of Dy f (3, −1, 2) and Dx f (3, −1, 2) are equal to
1 1 −1
1
−1
and Dx f (3, −1, 2) =
.
[Dy f (3, −1, 2)] =
2
1
3 1
So,
(c) Since
1 1 −1
1 0
1
0
Dg(3) = −
·
=−
=
.
1
2
1
3
−1
3
3
∂f
1
2
= −3,
(3, −1, 2) = det
det
1 −1
∂(x, y1 )
we may apply the implicit function to solve the equation f (x, y1 , y2) = 0
for x, y1 in terms of y2 .