MATH2000 TUTORIAL 6 Fall 2006 1. (a) Prove that if f and g are ...

MATH2000 TUTORIAL 6 Fall 2006
1. (a) Prove that if f and g are continuous functions defined on the same domain D in
R N (into R), prove that the map F : D → R 2 defined by F (x) = (f(x), g(x)) is
continuous.
(b) Prove that the functions a: R 2 → R and m: R 2 → R defined by a(x, y) = x+y
and m(x, y) = xy are continuous
(c) Use (a), (b) above, and the fact that continuity is preserved under composition,
to prove that if f, g are continuous functions on D, then so are f + g and fg.
2. (a) Let Dd = R × (R\{0}) ≡ {(x, y) ∈ R 2 : y = 0}. Prove that the function d on
Dd defined by d(x, y) = x/y is continuous. Use the method of the previous exercise
to deduce that if f and g are continuous functions defined on the same domain D,
and if g(x) = 0 for all x ∈ D, then the function f/g on D is continuous. (b) Use
the continuity of exponential and logarithm functions to prove that if f and u are
continuous functions with the same domain D in R N and if u(x) > 0 for all x ∈ D,
then the function u f is also continuous.
3. Prove that if f : D → R M is continuous, where D is a closed set in R N , then
Gf = {(x, y) ∈ R N × R M : x ∈ D, y = f(x)},
called the graph of f, is a closed set in R N × R M = R M+ N . (Simply put, the graph
of a continuous map (on a close set) is closed. The converse is not true. However,
there is a deep theorem in functional analysis called the closed graph theorem ...
) Hint: take a point p in the closure of Gf so that it is approachable by a sequence
in Gf and use continuity of f to prove that p is actually in Gf.
4. (a) Prove that if f : D → R is a continuous function defined on a bounded closed set
D in R N , then its range f(D) is also closed. (b) Prove that the above assertion fails
if the boundedness assumption on D is dropped. (Hint for (a): take a point in the
closure of f(D) and approach this point by a sequence in f(D). Hint for (b): consider
the function f on R defined by f(x) = x 2 /(x 2 + 1). Then 1 ∈ f(D) but 1 ∈ f(D). )

MATH2000 ANSWERS TO TUTORIAL 6 Fall 2006
1. (a) Take any point x0 ∈ D. Let ε > 0 be given. By the continuity of f and g, we
know that there exists δ > 0 such that |f(x) − f(x0)| < ε/2 and |g(x) − g(x0)| < ε/2
for all x ∈ D ∩ Bδ(x0). Consequently, for all x ∈ D ∩ Bδ(x0),
F (x) − F (x0) = (f(x), g(x)) − (f(x0), g(x0)) = (f(x)−f(x0)) 2 +(g(x)−g(x0) 2
≤ |f(x) − f(x0)| + |g(x) − g(x0)| < ε/2 + ε/2 = ε.
Hence F = (f, g) is continuous. (Here we have used an elementary inequality for
given nonnegative numbers a, b: √ a 2 + b 2 ≤ a + b. This can be verified as follows:
√ a 2 + b 2 ≤ √ a 2 + 2ab + b 2 = a + b.)
(b) Suppose (x0, y0) ∈ R 2 and ε > 0 are given. Let δ be a positive number to be
determined. Let (x, y) ∈ R 2 with (x, y) − (x0, y0) ≡ (x − x0) 2 + (x − x0) 2 < δ.
Then we have |x − x0|, |y − y0| < δ and hence
|a(x, y) − a(x0, y0)| = |(x + y) − (x0 + y0)| ≤ |x − x0| + |y − y0| < 2δ
So |a(x, y) − a(x0, y0)| < ε when we let δ = ε/2. Also, assuming δ < 1, we have
|m(x, y) − m(x0, y0)| = |xy − x0y0| = |(x − x0)y + x0(y − y0)|
≤ |x − x0|(|y − y0| + |y0|) + |x0||y − y0| < δ(1 + |y0|) + |x0|δ
So we have |m(x, y) − m(x0, y0)| < ε when we let δ = (1 + |x0| + |y0|) −1 ε.
(c) From the fact that continuity is preserved under composition, we see that the
functions f + g = a ◦ (f, g) and fg = m ◦ (f, g) are continuous.
2. (a) Take any point (x0, y0) ⊂ R 2 with y0 = 0. Suppose that ε > 0 is given. Set
δ > 0 to be determined. Take (x, y) ∈ R 2 with y = 0 such that (x, y)−(x0, y0) < δ.
Then |x − x0|, |y − y0| < δ and hence
|d(x, y) − d(x0, y0)| =
x x0
−
y y0
= |xy0 − x0y|
|y||y0|
= |(x − x0)y − x0(y − y0)|
|y||y0|
≤ |x − x0|(|y − y0| + |y0|) + |x0||y − y0|
(assuming |y − y0| < |y0|)
(|y0| − |y − y0|)|y0|
≤ 2δ|y0| + |x0|δ
(assuming |y − y0| ≤ |y0|/2)
(|y0|/2)|y0|
= Mδ, where M = 4(|x0| + |y0|)/|y0| 2 .

Notice that M ≥ 2|y0|/|y0| 2 = 2/|y0|. Now we set δ = M −1 min{ε, 1}. Then
|y − y0| < δ implies |y − y0| < M −1 ≤ |y0|/2. Thus (x, y) − (x0, y0) < δ implies
|d(x, y) − d(x0, y0)| < ε. The continuity of f/g follows from f/g = d ◦ (f, g) and the
argument used in the previous exercise.
(b) For (x, y) ∈ R 2 with y > 0, we have
h(x, y) ≡ y x = (e ln y ) x x(ln y)
= e
giving us h = exp ◦m ◦ (id, ln), where id stands for the identity function, m stands
for the multiplication function defined in the previous exercise. Thus h is a continuous
function. Hence u f = h ◦ (f, u) is continuous.
3. It is enough to check that the closure Gf of Gf is contained in Gf. Take a point
p = (x, y) in Gf where x ∈ R N and y ∈ R M . Then there is a sequence of
points {pn} in Gf converging to p. For each n, pn = (xn, f(xn)) for some
xn ∈ D. Now limn→ ∞ pn = p becomes limn→ ∞ (xn, f(xn)) = (x, y), which implies
limn→ ∞ xn = x and limn→ ∞ f(xn) = y. Since D is closed and xn ∈ D for all n,
the limit x of {xn} is also in D. The continuity of f gives limn→ ∞ f(xn) = f(x).
In view of limn→ ∞ f(xn) = y, we get y = f(x). Thus we have x ∈ D and
p = (x, y) = (x, f(x)), showing that p ∈ Df.
4. (a) It suffices to show f(D) ⊆ f(D). Take any point y0 in the closure f(D) of
f(D). Then there is a sequence {yn} in f(D) converges to y0. For each n, there
exists xn ∈ D such that yn = f(xn). Since D is bounded and closed, in view of the
Bolzano-Weierstrass theorem, the sequence {xn} in D has a convergent subsequence,
say {xnk } with limk→ ∞ xnk = x0 for some x0 ∈ D. By the continuity of f, we have
limk→ ∞ f(xnk ) = f(x0). On the other hand, limk→ ∞ f(xnk ) = limk→ ∞ ynk = y0. So
f(x0) = y0, showing that y0 ∈ f(D).
(b) Consider the function f on R defined by f(x) = x 2 /(x 2 + 1). Then
lim
n→ ∞
f(n) = lim
n→ ∞
n 2
n 2 + 1
= lim
n→ ∞
1
= 1,
1 + n−2 showing 1 ∈ f(R). But 1 ∈ f(R), because 1 ∈ f(R) would give a 2 /(1 + a 2 ) = 1
for some a ∈ R, or a 2 = a 2 + 1, which is impossible. Hence f(R) is not closed.