If V is the real vector space of all mapping from R to R.V1 = {f ∈ V: f(-x) = f(x)} and V2 = {f ∈ V: f(-x) = -f(x)}, then which one of the following is correct.

If V is the real vector space of all mapping from R to R.V1 = {f ∈ V: f(-x) = f(x)} and V2 = {f ∈ V: f(-x) = -f(x)}, then which one of the following is correct. Correct Answer both V₁ and V₂ are subspaces of V

Concept:

x, y ∈ S ⇒ x + y ∈ S,      ----(closed under addition)

x ∈ S ⇒ rx ∈ S      for all r ∈ R      ----(closed under multiplication)

Analysis:

V1 = {f ∈ V: f(-x) = -f(x)}

1) Take any two functions f, g ∈ V₁

f(-x) = -f(x) and g(-x) = -g(x);   ∀ x ∈ R

Then,

(f + g)(-x) = f(-x) + g(-x) = -f(x) - g(x) = - = -(f + g)(x)

That means, (f + g) is an odd function, ∀ x ∈ R 

(f + g) ∈ V₁

2) Take any function f ∈ V1 and c ∈ R

f(-x) = -f(x);   ∀ x ∈ R

Then,

(c f)(-x) = c f(-x) = c = -cf(x) = -(cf)(x)

That means, cf is an odd function, ∀ x ∈ R

cf ∈ V1

∴ V₁ is a subspace of V

Now, 

V₂ = {f ∈ V: f(-x) = f(x)}

1) Take any two functions f, g ∈ V₂

f(-x) = f(x) and g(-x) = g(x);   ∀ x ∈ R

Then,

(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)

That means, (f + g) is an even function, ∀ x ∈ R 

(f + g) ∈ V2

2) Take any function f ∈ V2 and c ∈ R

f(-x) = f(x);   ∀ x ∈ R

Then,

(c f)(-x) = c f(-x) = c = (cf)(x)

That means, cf is an even function, ∀ x ∈ R

cf ∈ V2

∴ V2 is a subspace of V