Subspace of C[-1,1]: f(-1)=0, f(1)=0

[Dustinsfl]

The set of all functions f in C[-1,1], f(-1)=0 and f(1)=0.

Nonempty since f(-1) = x^(2n) - 1 and f(1) = x^n - 1 ϵ C[-1,1] where n ϵ ℤ, n ≥ 0

α·x^(2n) - α = α (x^(2n) - 1) = α·0 = 0 and α·x^n - α = α (x^n - 1) = α·0 = 0

x^(2n) - 1 + x^n - 1 = (x^(2n) - 1) + (x^n - 1) = 0 + 0 = 0

Based on this example C is a subspace; however, I can't think of how to do a general solution since there are more equations that satisfy the requirements. For instance, x + 1 when x = -1.

 

[Dick]

You are making some strange statements there. x+1 is NOT is C[-1,1]. An f in C[-1,1] need to satisfy BOTH f(-1)=0 and f(1)=0. That rules out x^n-1 if n is odd as well. x^(2n)-1 is ok. But you aren't supposed to be worrying about specific functions. You are supposed to show if f(x) and g(x) are in C[-1,1] then f(x)+g(x) and a*f(x) are in C[-1,1]. You don't have to write out some special form for f(x) and g(x) to do that.