Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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)?