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Functions to and from R

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Functions to and from R

n

24/05

Let F : D → Rm be a function (mapping) where D ⊆ Rn. Examples:

γ: [0, 2π ] → R3, t , (cos t, sin t, t)

f : R3 \ {0} → R, (x, y, z) , x2+y12+z2 = r12

fA: Rn → Rm, X , AX, A ∈ Rm,n (a linear map) Notation. We shall write v = (x, y, z) ∈ R3

but x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn Definition. The distance between x, y ∈ Rn is

kx − yk = s n

P

k=1

(xkyk)2.

This is the usual distance for n = 2, 3 by Pythagoras:

dist (x, y), (x0, y0)2

= (x − x0)2 +(y − y0)2, and generalizes the absolute value for n = 1:

dist(5, −2) = |5 − (−2)| = 7.

Let F : RnD → Rm, so

F (x) = (F1(x), . . . , Fm(x))

= (F1(x1, . . . , xn), . . . , Fm(x1, . . . , xn)).

We want to say what it means for F to be continuous. The definition is the same as that for an ordinary function, and is best explained using limits:

Let a ∈ Rn. We say that lim

x→a F (x) exists if there exists ` ∈ Rm such that “F (x) approaches ` whenever x approaches a.”

This means: given 101 , there exists δ > 0 such that 0 < kx − ak < δ ⇒ kF (x) − `k < 101 .

(It is important to note that we are not allowed to test x = a).

Now replace 101 by an arbitrary ε > 0!

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It may be that a ∉ D so that F(a) is not defined, but if it is:

Definition. F : RnD → Rm is continuous at a ∈ D if limx→aF (x) exists and equals F (a).

Equivalently, given ε > 0, ∃δ > 0 such that kx − ak < δ ⇒ kF (x) − F (a)k < ε.

Since kF (x)k2=P |Fk(x)|2, F will be continuous if and only if the components F1, . . . , Fm are all continuous. Thus, we only considered examples with m = 1 and n = 2:

Let D = R2 \ {(0, 0)}. Then we showed that f (x, y) =x

x2+y2 = cosθ is not continuous at a = (0, 0).

G(x, y) =

0 if (x, y) = (0, 0)

xy

x2+y2 otherwise is continous at (0, 0).

One can appreciate the definition of continuity with the

Proposition. Any linear mapping f : Rn → Rm is continuous at every point of its domain D = Rn!

Proof. Let A ∈ Rm,n be the matrix associated to f (relative to the canonical bases) and let x, a ∈ Rn,1. Then

kf (x) − f (a)k = kAx − Aak = kA(x − a)k.

It can be shown (not easily even if m = n = 2) that the last term is less than or equal to kAkkx − ak where kAk =q

P aij2. Anyway,

kx − ak < 10kAk1 ⇒ kf (x) − f (a)k < 101

(we can assume that A is not the zero matrix so kAk 6= 0), and continuity follows replacing 101 by an arbitrary ε > 0.

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