Download CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES 1

CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES
1. Definitions
n
1.1. Limit. Let f : R → R
(y1 , . . . , ym ) ∈ Rm . Then
m
some function, x0 = (x1 , . . . , xn ) ∈ Rn and y0 =
lim f (x) = y0
x→x0
if and only for x ”close to” x0 , f (x) is ”close to” y0 . In other words, given > 0,
there exists δ depending on and x0 (so we write δ = δ(, x0 )) such that
dRn (x, x0 ) < δ ⇒ dRm (f (x), y0 ) < where the distance on the left is taken in Rn , whereas the one on the right-hand
side is taken in the target space Rm . We will drop the subscript Rn from d, and
we will often write kx − x0 k instead of d(x, x0 ). We will also drop the bold-face
notation.
Note: a function f : Rn → Rm is clearly given by a row vector f = (f1 , . . . , fm )
where fi ’s are the components of f . In fact, fi = pi ◦ f (see below). Then
limx→x0 f (x) = y0 if and only if limx→x0 fi (x) = yi , i = 1, . . . , m.
1.2. Continuity. A function f : Rn → Rm is continuous at x0 ∈ Rn iff
lim f (x) = f (x0 )
x→x0
We say that f is continuous (everywhere) if it is continuous at every point of the
domain Rn .
Note: if f = (f1 , . . . , fm ), where f1 , . . . , fm are the components of f , then f is
continuous iff f1 , . . . , fm are continuous, as functions : Rn → R.
2. Tools
Operations with limits: addition, subtraction, multiplication, division (when possible), etc.
Components. If f : Rn → Rm , f = (f1 , . . . , fm ), then limx→x0 f (x) = y0 =
(y1 , . . . , ym ) iff limx→x0 fi (x) = yi , for every i = 1, . . . , m.
Projections. The functions pi : Rn → R, pi (x1 , . . . , xn ) = xi are continuous, and
this can be easily checked with the (, δ)- definition of continuity.
Composition of functions. If f : Rn → Rm is continuous and g : Rm → RN is
continuous, then g ◦ f : Rn → RN is continuous, where g ◦ f (x) := g(f (x)).
f
g
Rn −−−−→ Rm −−−−→ RN
1
2
CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES
3. Examples
1. f : R3 → R, f (x, y, z) = x sin(z).
Then f = p1 · (sin ◦p3 ) = continuous.
2. f : R3 → R2 , f (x, y, z) = (xy, x + y + z 2 ).
• f1 (x, y) = xy = p1 · p2 = cont.
• f2 (x, y) = x + y + z 2 = p1 + p2 + (p3 )2 = cont.
Therefore f = (f1 , f2 ) is continuous.
( x2
√ 2 2 , (x, y) 6= (0, 0)
2
x +y
3. f : R → R, f (x, y) =
Proof that f is continuous.
0,
(x, y) = (0, 0)
On R2 − (0, 0), f is continuous since it is a ratio of continuous functions (with
p2
non-vanishing denominator), namely f = √·◦(p21+p2 ) = continuous.
1
2
NB: To make sure we understand why is the denominator continuous, we can write
the denominator as the composition of functions
√
p2 +p2
·
2
R2 −−1−−→
[0, ∞) −−−−→ R
We are thus left to show that f is continuous at (0, 0), in other words to prove that
x2
lim
(x,y)→(0,0)
p
x2 + y 2
=0
Squeeze:
x2
≤ √ = |x|
x2
x2 + y 2
However, as (x, y) → (0, 0) necessarily x → 0 and hence |x| → 0, therefore by the
2
Pinching Lemma √ x2 2 → 0 as (x, y) → 0.
0≤ p
x2
x +y
Different way of saying it ((, δ)-proof). Note that
p
x2
≤ |x| ≤ d((x, y), (0, 0)) ≤ ⇔ x2 + y 2 ≤ ⇒ |x| ≤ ⇒ p
x2 + y 2
(
xy
(x, y) 6= (0, 0)
2
2,
2
4. f : R → R, f (x, y) = x +y
.
0,
(x, y) = (0, 0)
Proof that f is not continuous at (0, 0). Consider the followin sequences of points
in R2 :
1
xk = ( , 0), k ≥ 1
k
1 1
yk = ( , ), k ≥ 1
k k
Then xk → (0, 0) and yk → (0, 0) as k → ∞, and yet
f (xk ) = 0 → 0, k → ∞
1
1
f (yk ) = → , k → ∞
2
2
which shows that the limit
lim f (x)
x→(0,0)
does not exist. In particular, the function f is not continuous at (0, 0).
CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES
5. Let
3
(
x sin(1/y), y 6= 0
g(x, y) =
0,
y=0
Determine the points (x, y) ∈ R2 where g is continuous.
6. Tricky example. Let f : R2 → R defined by
(
xy 2
(x, y) 6= (0, 0)
2 +y 4 ,
x
f (x, y) =
0,
(x, y) = (0, 0)
Show that limk→∞ f (xk ) = 0, for any sequence xk → (0, 0) such that xk , k ≥ 1 is
on a fixed line passing through (0, 0). Is f continuous at (0, 0)?