MAS221 Analysis, Semester 2 Solutions
Sarah Whitehouse
Chapter 6 and 7 Problems
Chapter 6 Problems: Functions of several variables
1. Let f (x, y) = (u, v). Then
u = −y,
v = x2 + y,
y = −u,
x2 = u + v.
that is to say
Now 0 ≤ x ≤ 2, so 0 ≤ x2 ≤ 4, and 0 ≤ y ≤ 1, so
0 ≤ −u ≤ 1,
0 ≤ u + v ≤ 4,
−1 ≤ u ≤ 0,
0 ≤ u + v ≤ 4.
that is to say
This is the quadrilateral enclosed by the lines u = −1, u = 0, u + v = 0,
and u + v = 4.
2. (a) f −1 (1) is the set of points lying on a circle of radius 1 around the
origin, f −1 (0) is the point (0, 0) and f −1 (−1) is the empty set.
(b) The pre-image f −1 [(0, 1]] is a disk of radius 1 around the point (0, 0),
including the boundary points, but missing the point (0, 0).
3. (a) Let (x, y) ∈ R2 . Let (xn , yn ) be a sequence in R2 converging to (x, y).
Then xn → x and yn → y as n → ∞.
By the algebra of limits, xn yn → xy as n → ∞. Hence f (xn , yn ) →
f (x, y) as n → ∞.
It follows that the function f is continuous.
(b) Let (x, y) ∈ R2 . Let (xn , yn ) be a sequence in R2 converging to (x, y).
Then, since the functions f and g are continuous, f (xn , yn ) → f (x, y)
and g(xn , yn ) → g(x, y) as n → ∞.
By the algebra of limits, f (xn , yn )g(xn , yn ) → f (x, y)g(x, y) as n →
∞.
It follows that the product function f g is continuous.
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4. The functions (x, y) 7→ xy, s 7→ sin s and t 7→ et are continuous. The
composite of continuous maps is continuous. Hence the function f is
continuous.
5. Define g : [0, 1] → R by g(t) = f (ty + (1 − t)x). Then g is continuous,
since f is. Observe that g(0) = f (x) and g(1) = f (y). Then by the
intermediate value theorem, we have t ∈ [0, 1] such that g(t) = c.
Let z = ty + (1 − t)x. Then we have g(t) = f (z) = c.
6. By the Bolzano-Weierstrass Theorem, every bounded sequence in R2 or R
has a convergent subsequence.
Let f : [a, b] × [c, d] → R be continuous. Let (yn ) be a sequence in the
image of f . Write yn = f (xn ).
Then by definition of the domain of f , the sequence (xn ) is bounded.
Hence it has a convergent subsequence (xnk ). Since [a, b] × [c, d] is closed,
the limit of this subsequence lies in [a, b] × [c, d]. Since f is continuous,
the subsequence ynk = f (xnk ) converges.
Thus every sequence in the image of f has a convergent subsequence, and
the limit of that subsequence also lies in the image.
Suppose the function f is not bounded. Then there is a point yn in the
image of f such that |yn | ≥ n. But this sequence has no convergent
subsequence, which contradicts the above. Thus f is bounded.
Let M = sup{f (x) | x ∈ [a, b] × [c, d]}. Then we have an element yn in
the image of f such that M ≥ yn > M − n1 . It follows that yn → M as
n → ∞.
Set yn = f (xn ). Then the sequence (xn ) has a convergent subsequence;
call its limit x. By continuity of the function f , we have f (x) = M , and
f attains its maximum.
Similarly, f attains its minimum.
7. (a) The functions g and h are constant functions, with value 0, and are
therefore continuous.
(b) Let t > 0. Observe that
f (t, t) =
1
t2
= .
2
2
t +t
2
Hence f (t, t) → 21 6= 0 = f (0, 0) as t → 0. Therefore the function f
cannot be continuous at (0, 0).
Chapter 7 Problem: Applications II
1. (a) Continuity is standard at x, for 0 < x < 1. We need to check
continuity at x = 0 and at x = 1.
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Let 0 < x < 1, so
ft (x) =
xt − 1
.
ln x
We have xt − 1 → −1 6= 0 as x → 0, and ln x → −∞ as x → 0. So
clearly
xt − 1
→ 0 = ft (0)
ln x
as x → 0, and ft is continuous at 0.
On the other hand,
xt − 1 → 1 − 1 = 0
and
ln x → 0
as x → 1, so by L’Hôpital’s rule, we can differentiate the top and
bottom:
xt − 1
txt−1
lim
= lim
= t = ft (1)
x→1 ln x
x→1 1/x
so ft is continuous at 1.
(b) We say the sequence (fn ) converges uniformly to a function
f : [0, 1] → R when for all ε > 0, we have N ∈ N such that
|fn (t) − f (t)| < ε for all n ≥ N and all t ∈ [0, 1].
In Chapter 3, Q7 we saw that the sequence of functions (f1/n ) converges uniformly to the zero function.
(c) Since we can assume the function ft is increasing, we have that
|ft (x)| ≤ |ft (1)| = t
for all t ∈ [0, 1].
Let (tn ) be a sequence of positive real numbers with limit 0. Then
|ftn (x)| ≤ tn for all x, and tn → 0 as n → ∞. Therefore (ftn )
converges uniformly to the zero function.
The result now follows by the uniform convergence theorem for integrals.
∂h
∂t
(d) It suffices that h and
are continuous on the domain [0, 1] × R.
(e) Let
Z
I=
0
1
xt − 1
dx =
ln x
Z
0
1
et ln x − 1
dx.
ln x
Differentiating under the integral sign, as above
Z 1 t ln x
Z 1
dI
e
ln x
=
dx =
et ln x dx
dt
ln x
0
0
t+1 x=1
Z 1
x
1
=
xt dt =
=
.
t + 1 x=0
t+1
0
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So, integrating the right hand side
I = ln(t + 1) + C
where C is constant.
By the above, if we let t → 0, the right-hand side converges to C,
and the left-hand side converges to 0. Therefore C = 0, and
I = ln(t + 1)
as required.
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