### Functions

### Definition

Let A and B be two subsets of ℝ

A

A

**relation**from A to B is a rule which assigns a member of B to each member of A.### Definition

Let A and B are two subsets of ℝ

A

A

**function**from A to B is a rule which assigns a unique member of B of each number of A.
In the example above, 5 being assigned to both 2 & 6. Hence the relation in that example is not a function.

Here is an example for a function;

Here is an example for a function;

### Definition

Let f: A → B a function.

If f(x

We say that f is an one to one function

In the previous example f

∴ f

If f(x

_{1}) = f(x_{2}) ⇒ x_{1}= x_{2}We say that f is an one to one function

In the previous example f

_{o}(1) = f_{o}(9) but 1 ≠ 9∴ f

_{o}is not a one to one function.### Definition

Let f: A → B be a function

If ∀y ∈ B, ∃x ∈ A s.t. f(x) = y

We say that f is an onto function

In the previous example ∃!x s.t. f

∴ f

If ∀y ∈ B, ∃x ∈ A s.t. f(x) = y

We say that f is an onto function

In the previous example ∃!x s.t. f

_{o}(x) = 4∴ f

_{o}is not an onto function### Definition

Let f: A → B be a function.

If f is both one to one and onto, we say that f is a bijection

In the previous example f

∴ f

If f is both one to one and onto, we say that f is a bijection

In the previous example f

_{o}is neither one to one nor onto.∴ f

_{o}is far from being bijection### Definition

Let f: A → B be a function

- A is called the Domain of A We write D(f) = A
- The following set is called the range of f R(f) = {y | y ∈ B & y = f(x) for some x ∈ A}

### Definition

Let f: A → B be a function

f(x) = 2x + 7

x

⇒ f(x

∴ f is a function.

x

∴ f is a strictly increasing function.

- If x
_{1}< x_{2}⇒ f(x_{1}) ≤ f(x_{2}), f is an**increasing function** - If x
_{1}< x_{2}⇒ f(x_{1}) < f(x_{2}), f is an**strictly increasing function** - If x
_{1}< x_{2}⇒ f(x_{1}) ≥ f(x_{2}), f is an**decreasing** - If x
_{1}< x_{2}⇒ f(x_{1}) > f(x_{2}), f is an**strictly decreasing function**

f(x) = 2x + 7

x

_{1}= x_{2}⇒ 2x_{1}= 2x_{2}⇒ 2x_{1}+ 7 = 2x_{2}+ 7⇒ f(x

_{1}) = f(x_{2})∴ f is a function.

x

_{1}< x_{2}⇒ 2x_{1}< 2x_{2}⇒ 2x_{1}+ 7 < 2x_{2}+ 7∴ f is a strictly increasing function.

### Definition

Let x ∈ R

The greatest integer less than or equal to x is denoted by [x]

Eg: [7.3] = 7

The greatest integer less than or equal to x is denoted by [x]

Eg: [7.3] = 7

### Definition

Let f: A → B be a function

The inverse relation f

f

in general f

The inverse relation f

^{-1}R(f) → A is defined in the following manner.f

^{-1}(y) = x where y = f(x) for some x ∈ Ain general f

^{-1}is not a function### Theorem

Let f: A → B be a function.

Then f

Then f

^{-1}is a function ⇔ f is one to one### Proof

Assume f

Write f(x

f(x

f(x

x

∴ f is one to one.

Assume f is one to one

Let y ∈ R(f) suppose f

f

f

f(x

∴ x

∴ f

^{-1}is a function. Let f(x_{1}) = f(x_{2})Write f(x

_{1}) = f(x_{2}) = yf(x

_{1}) = y ⇒ f^{-1}(y) = x_{1}f(x

_{2}) = y ⇒ f^{-1}(y) = x_{2}x

_{1}= x_{2}because f^{-1}is a function∴ f is one to one.

Assume f is one to one

Let y ∈ R(f) suppose f

^{-1}(y) = x_{1}& f^{-1}(y) = x_{2}f

^{-1}(y) = x_{1}⇒ f(x_{1}) = yf

^{-1}(y) = x_{2}⇒ f(x_{2}) = yf(x

_{1}) = f(x_{2}) & f is one to one∴ x

_{1}= x_{2}∴ f

^{-1}is a function### Definition

Let f: A → B & g: A → B be two functions

- f + g: A → B (f + g)(x) = f(x) + g(x)
- f - g: A → B (f - g)(x) = f(x) - g(x)
- rf: A → B (rf)(x) = rf(x)
- fg: A → B (fg)(x) = f(x).g(x)
- f/g: A → B (

^{f}/

_{g})(x) =

^{f(x)}/

_{g(x)}(g(x) ≠ 0)

### Composition

### Definition

Let f: A → B & g: B → C be two functions. The composite function (g

g

(g

eg:

g

(g

_{o}f) is defined asg

_{o}f: A → C(g

_{o}f)(x) = g(f(x))eg:

f: R → R | g: R[-1, 1] |

f(x) = x_{2} – 3x + 2 |
g(x) = sin x |

g

_{o}f: R → [-1, 1](g

_{o}f)(x) = g(f(x)) = g(x_{2}– 3x + 2) = sin(x_{2}– 3x + 2)
A function f: R → R is said to be an additive function if the following condition satisfied.

f(x + y) = f(x) + f(y) ∀x, y ∈ ℝ

0 + 0 = 0 ⇒ f(0 + 0) = f(0)

f(0) + f(0) = f(0) because f is additive.

∴ f(0) = 0

Let m ∈ ℤ

Claim: f(mx) = mf(x) ∀x ∈ ℝ

It is clearly true for m = 1

Assume the claim is true for m = k

f{(k + 1)x} = f(kx + x) = f(kx) + f(x) [because f is additive]

= kf(x) + f(x)

=(k + 1)f(x)

∴ The claim is true for m = k + 1

∴ By induction, claim is true ∀m ∈ ℤ

Let m ∈ ℤ

Now m + n = 0 ⇒ (m + n)x = 0.x = 0

∴ f(mx + nx) = f(0) = 0 [by part I]

∴ f(mx) + f(nx) = 0 [because f is additive]

f(mx) + f(nx) = 0

f(mx) = -nf(x) = mf(x)

So we have proved that f(mx) = mf(x) ∀m ∈ ℤ

Let r ∈ ℚ

Write r =

Let x ∈ ℝ x = n(

f(x) = f[n(

f(rx) = f(

= m.

f(x + y) = f(x) + f(y) ∀x, y ∈ ℝ

*Prove the following fact for an additive function.*- f(0) = 0
- f(rx) = rf(x) ∀r ∈ ℚ, ∀x ∈ ℝ

**Proof of i.**0 + 0 = 0 ⇒ f(0 + 0) = f(0)

f(0) + f(0) = f(0) because f is additive.

∴ f(0) = 0

**Proof of ii.**Let m ∈ ℤ

^{+}Claim: f(mx) = mf(x) ∀x ∈ ℝ

It is clearly true for m = 1

Assume the claim is true for m = k

f{(k + 1)x} = f(kx + x) = f(kx) + f(x) [because f is additive]

= kf(x) + f(x)

=(k + 1)f(x)

∴ The claim is true for m = k + 1

∴ By induction, claim is true ∀m ∈ ℤ

^{+}Let m ∈ ℤ

^{-}Now m + n = 0 ⇒ (m + n)x = 0.x = 0

∴ f(mx + nx) = f(0) = 0 [by part I]

∴ f(mx) + f(nx) = 0 [because f is additive]

f(mx) + f(nx) = 0

f(mx) = -nf(x) = mf(x)

So we have proved that f(mx) = mf(x) ∀m ∈ ℤ

Let r ∈ ℚ

Write r =

^{m}/_{n}with m ∈ ℤ, n ∈ ℤ^{+}Let x ∈ ℝ x = n(

^{1}/_{n}x)f(x) = f[n(

^{1}/_{n}x)] = nf(^{1}/_{n}x)^{1}/_{n}f(x) = f(^{1}/_{n}x)f(rx) = f(

^{m}/_{n}x) = f[m(^{1}/_{n}x)] = mf(^{1}/_{n}x)= m.

^{1}/_{n}f(x) =^{m}/_{n}f(x) = rf(x)
## Comments

## Post a Comment