Friday, March 10, 2006

Combining Continuous Functions

Continuous functions can be combined to form new, more complicated continued functions. In today's blog, I will review some basic lemmas on how functions built from continued functions are themselves continued functions. Finally, I will show how these lemmas can be used to establish the minimum property of continuous functions over a closed interval.

If you are not familiar with continuous functions, then should review here.

Lemma 1: Constant Law

f(x) = C where C is a constant is a continuous function

(1) Let δ = 1

(2) Let x,c be any two values in the domain of f(x).

(3) We know that f(x) - f(c) = C - C = 0

(4) So, it is also true, if x-c is between and , for any nonzero value ε, that f(x) - f(c) is between and [Since 0 is between and ]

(5) So, by the definition of continuous functions (see here), f(x)=C is a continuous function.

QED

Lemma 2: Addition Law

if f(x) and g(x) are continuous functions, then f(x) + g(x) is a continuous function.

(1) Let ε be any nonzero value.

(2) Let c be any point on the domain of f(x) + g(x).

(3) Since f(x) is continuous, we know that there exists δ1 such that:
if x - c is between 1 and 1, then f(x) - f(c) is between -ε/2 and +ε/2

(4) Since g(x) is continuous, we know that there exists δ2 such that:
if x - c is between 2 and 2, then g(x) - g(c) is between -ε/2 and +ε/2

(5) Let δ = min(δ12)

(6) Now, if x - c is between and , then:

(a) f(x) - f(c) is between -ε/2 and +ε/2

(b) g(x) - g(c) is between - ε/2 and + ε/2

(c) f(x) + g(x) - f(c) - g(c) is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2)

(7) So, f(x) + g(x) is a continous function.

QED

Lemma 3: Multiplication Law
if f(x) and g(x) are continuous functions, then f(x)*g(x) is a continuous function.

(1) Let ε be any nonzero value.

(2) Let c be any point on the domain of f(x)*g(x).

(3) Since f(x) is continuous, we know that there exists δ1 such that:
if x - c is between 1 and 1, then f(x) - f(c) is between and

So there exists a constant A = max(absolute[ε + f(c)], absolute[-ε + f(c)]) such that:
if x - c is between 1 and 1, then f(x) is between -A and +A.

(4) Let B = g(c).

(5) We also know that there exists δ2 such that:
if x - c is between 2 and 2, then f(x) - f(c) is between -ε/(2B) and ε/(2B).

(6) And there exists δ3 such that:
if x - c is between 3 and 3, then g(x) - g(c) is between -ε/(2A) and ε/(2A)

(7) Let δ = min(δ123)

(8) Now, if x - c is between and , then:

(a) f(x) - f(c) is between -ε/(2B) and +ε/(2B)

(b) g(x) - g(c) is between - ε/(2A) and + ε/(2A)

(c) g(c)[f(x) - f(c)] is between (B)[-ε/(2B)] and (B)[+ε/(2B)] which is between -ε/2 and ε/2.

(d) f(x)[g(x) - g(c)] is between (A)[-ε/(2A)] and (A)[+ε/(2A)] which is between -ε/2 and ε/2.

(e) If we add (c) + (d), we get:
f(x)g(c) - f(c)g(c) + f(x)g(x) - f(x)g(c) = f(x)g(x) - f(c)g(c)

(f) So, f(x)g(x) - f(c)g(c) is between (-ε/2 + -ε/2) and (+ε/2 + +ε/2)

(7) So, f(x)g(x) is a continous function.

QED

Lemma 4: Differentiability implies continuity

If f is differentiable at a, then f is continuous at a

Proof:
(1) Since f is differentiable at a, there exists a value c such that f'(a) = c [See here for definition of derivative]

(2) Let ε be any nonzero value

(3) Let δ = Δx

(4) By the definition of the derivative (see here), we know that:
if x - a is between and δ, then f(x) = c so that f(x) - f(c) = 0 which is between ε and

(5) Therefore, by the definition of continuity at a point (see here), f(x) is continuous at point a.

QED

Theorem: Minimum Value Property of Continuous Functions

If a function f is continuous on a closed interval [a,b], then there exists a number c in [a,b] such that f(x) ≥ f(c) for all x in [a,b]

(1) We can construct a continuous function g(x) such that g(x) = (-1)f(x).

We know that g(x) is a continuous function since it is the product of a constant function h(x)=-1 and f(x). [See Lemma 1and Lemma 3 above]

(2) Now, g(x) has a maximum point on the interval [a,b] (from the Maximum Value Property of Continuous Functions, see here)

(3) But since g(x) = -f(x), this corresponds to a minimum point for f(x) since:

(a) Let c be the maximum for g(x)

(b) g(x)g(c) for all x in [a,b]

(c) But then:
-g(x) ≥ -g(c) for all x in [a,b]

(d) Which means that:
f(x) ≥ f(c) for all x in [a,b]

QED

References

3 comments :

John said...

Correction: - g(c) not + g(c) for the additive properties of continuity.

Larry Freeman said...

Hi John,

Thanks for noticing! I've fixed the typo.

Cheers,

-Larry

Garbagemin Franklin said...

I think there's something wrong with your proof of Lemma 4:

***

(1) Since f is differentiable at a, there exists a value c such that f'(a) = c
...

(4) By the definition of the derivative (see here), we know that:
if x - a is between -δ and δ, then f(x) = c ...

***

Unless I'm missing something, this is equivalent to the following (false) claim:

If f is differentiable at a with f'(a)=c, then there is an interval [a-δ, a+δ] on which f(x) = c.