We described why the set of all realvalued functions defined

We described why the set of all real-valued functions defined on the interval [a, b] is a vector space. The set of all continuous functions on [a, b], which we sometimes write as C[a, b] is a subspace. (a) What facts about continuous functions from calculus prove that C[a, b] is in fact a subspace? (b) Show that {f elementof C[a, b]: f(a) = f(b)} is a subspace of C[a, b].

Solution

Let V be the vector space of all real valued functions on [a,b] and

W be the set of all continuous functions on [a,b].

Clearly W is a subset of V.

(a) W is a subspace of V.

Proof: Let f and g be in V.

we know that the sum of two continuous functions is continuous.

So (f+g) , defined by (f+g)(x) =f(x) +g(x) is continuous on [a,b], as f and g are so.

Thus W is closed under (vector) addition.

Next, let f be in W, and c be a scalar-any real number.

f being continuous on [a,b], the function (cf) defined by (cf)(x)=cf(x) (this is the product two continuous functions, one of them constant and the other the function g) is continuous.

Thus W is closed under scalar multiplication.

These two facts prove that W is a subspace of V. (Note: The function 0 belongs to W--being a constant function, it is continuous)

(b) Now let U ={f: f continuous on [a,b] and f(a) =f(b)}.

To show that U is a subspace of W.=C[a,b].

1) U is additively closed.

If f and g belong to U, then f+g is clearly continuous on [a,b].

also (f+g)(a) =f(a)+g(a)

                    =f(b)+g(b) (as f and g belong to U)

                    = (f+g)(b).implies the sum f+g belongs to U

So U is additively closed.

(Note :the 0 function clearly belongs to U, as 0(a)=0(b) =0)

Let f be in U and c be any scalar.

(cf)(a) = c f(a)=cf(b) = (cf)(b) implies that U is closed under scalar multiplication.

Hence U is a subspace of W.