Vector space questions

Question:

If X is the set of all real valued continuous (differentiable or integrable) functions defined on some interval [ 0,1]. Then to show that X is a vector space over the field R with addition and scalar multiplication defined as follows: 

(𝔣 + g)x = 𝔣(x) + g(x) ∀ 𝔣,g ∈ X .......(1) 

(α 𝔣) x = α 𝔣(x) ∀ α ∈ R ,𝔣 ∈ X, x ∈ [ 0, 1] 

Solution. To show X be a vector space, we have to verify all the postulates of vector space. Let us verify ono by one. 

V₁ : (X, +) is an abelian group. 

Closure axiom for scalar multiplication : 

For α ∈ E R . 𝔣 ∈ X ⟹ (α𝔣)(x) = α𝔣(x) ∈ X 

⟹ α𝔣 ∈ X 

V₂: For any α ∈ R: 𝔣,g ∈ X, we have

[ α (𝔣 + g) ](x) = α [ (𝔣 + g) ](x) ∀ x ∈ [ 0, 1] 

= α [(𝔣(x)+ g(x) ]

= α (𝔣(x)+ αg(x) 

= (α𝔣)(x)+ (αg)(x)

= (α𝔣+ αg)(x)  ∀ x ∈ [ 0, 1]

Hence α (𝔣 + g) = α𝔣+ αg

V₃: For α, β ∈ R, 𝔣 ∈ X, we have

(α + β) 𝔣(x) = (α + β) 𝔣(x) ∀ x ∈ [ 0, 1]

= α𝔣(x) + β𝔣(x) as elements of R satisfy distributive law

= (α𝔣)(x) + (β𝔣)(x)

= (α𝔣 + (β𝔣)(x) ∀ x ∈ [ 0, 1]

Hence  (α + β) 𝔣 =  α𝔣 + β𝔣

V₄: For any α, β ∈ R, 𝔣 ∈ X, we have

[{(αβ) 𝔣}](x) = (α β) 𝔣(x)

= α [β 𝔣(x)] ( as elements of R are associative for multiplication)

= α [(β 𝔣)(x)]

= α [(β 𝔣)](x) ∀ x ∈ [ 0, 1]

Hence (αβ) 𝔣 = α (β 𝔣)

V₅: As 1 ∈ R, so for any 𝔣 ∈ X, we have

(1𝔣)x = 1𝔣(x) = 𝔣(x) ∀ x ∈ [ 0, 1]

( as 1 ∈ R is the multiplicative identity)

Hence 1𝔣 = 𝔣

Thus all postulate of vector space are satisfied showing that X(R) is a vector space.

Question. 

Let S be set of all ordered pairs of real numbers. Define sums and scalar multiplication of pairs as follows : 
(x₁ , y₁) + (x₂ , y₂) = (x₁ + x₂ y₁ + y₂) 
and c (x₁ , y₁) = (cx₁ , y₁) 
Show that S is not a vector space.

Solution. 

Let a, b are any two real numbers and x = (x₁ , y₁) ∈ S. Then 

(a + b)x = (a + b)(x₁ , y₁) = ((a + b) x₁ , y₁)

 = (ax₁ + bx₁ ,y₁ ) 

and ax + bx = a(x₁ , y₁) + b(x₁ ,y₁ )

= (ax₁ , y₁) + (bx₁ ,y₁ )

= (ax₁ + bx₁, y₁ + y₁ )

= (ax₁ + bx₁, 2y₁ )

(a + b)x ≠ ax + bx ∀ x ∈ S

∴ Hence S is not a vector space.