Question

verify the assertion. (Subspace example)

1) The set of continuous real-valued functions on the interval [0,1] is a subspace of R^[0,1]

This is from Linear Algebra Done Right- Sheldon Axler 3rd edition.

I don't understand why the solution uses a integral.

Answer 1

First note that

R^^[0,1] = { f | f:[0,1]→R }

and let

S={ f| f:[0,1]→R , such that f is continuous}

According to defination you need to show that the set of continuous functions is non empty and that it is closed under the operations of R^[0,1]

(And by the way note that solution does not need any integral to prove it u might have seen another exercise in but tha is different and here we have different one )

now coming to main point

take zero function say

f₀ :[0,1]→R such that

f₀(x)=0 , so clearly it is real valued continuous function

so f₀ belongs to S

now take two functions

say f and g belonging to S

you need to show that

f+g belongs to S

and we know that sum of two real valued continuous functions from [0,1] to R is again real valued continuous

so you can say

f+g belongs to S

so closed under addition

now you need check that it is closed under scalar multiplication

for that

let a€ R be any real number (scalar)

and f be continuous real valued function I.e, f belongs to S

now af is also continuous (real valued)

therefore all conditions are satisfied now

and you can say S is a subspace of R^[0,1]

that is Set of continuous real valued functions in [0,1] is a subspace of R^[0,1]