13.1. COMBINATIONS OF CONTINUOUS FUNCTIONS 237

≤ 2

|g(x)|2[| f (x)g(y)− f (y)g(y)+ f (y)g(y)− f (y)g(x)|]

≤ 2

|g(x)|2[|g(y)| | f (x)− f (y)|+ | f (y)| |g(y)−g(x)|]

≤ 2

|g(x)|2

[32|g (x)| | f (x)− f (y)|+(1+ | f (x)|) |g(y)−g(x)|

]≤ 2

|g(x)|2(1+2 | f (x)|+2 |g(x)|) [| f (x)− f (y)|+ |g(y)−g(x)|]

≡M [| f (x)− f (y)|+ |g(y)−g(x)|]

whereM ≡ 2

|g(x)|2(1+2 | f (x)|+2 |g(x)|)

Now let δ 2 be such that if |x−y|< δ 2, then

| f (x)− f (y)|< ε

2M−1

and let δ 3 be such that if |x−y|< δ 3, then

|g(y)−g(x)|< ε

2M−1.

Then if 0 < δ ≤min(δ 0,δ 1,δ 2,δ 3), and |x−y|< δ , everything holds and∣∣∣∣ f (x)g(x)

− f (y)g(y)

∣∣∣∣≤M [| f (x)− f (y)|+ |g(y)−g(x)|]

< M[

ε

2M−1 +

ε

2M−1

]= ε.

This completes the proof of the second part of (2). Note that in these proofs no effort ismade to find some sort of “best” δ . The problem is one which has a yes or a no answer.Either it is or it is not continuous.

Now begin on (3). If f is continuous at x, f (x) ∈ D(g)⊆ Rp, and g is continuous atf (x) , then g ◦f is continuous at x. Let ε > 0 be given. Then there exists η > 0 such thatif |y−f (x)| < η and y ∈ D(g), it follows that |g (y)−g (f (x))| < ε . It follows fromcontinuity of f at x that there exists δ > 0 such that if |x−z| < δ and z ∈ D(f), then|f (z)−f (x)|< η . Then if |x−z|< δ and z ∈D(g ◦f)⊆D(f), all the above hold andso

|g (f (z))−g (f (x))|< ε.

This proves part (3).Part (4) says: If f = ( f1, · · · , fq) : D(f)→Rq, then f is continuous if and only if each

fk is a continuous real valued function. Then

| fk (x)− fk (y)| ≤ |f (x)−f (y)| ≡

(q

∑i=1| fi (x)− fi (y)|2

)1/2

≤q

∑i=1| fi (x)− fi (y)| . (13.1)