Proving a set of derivatives to be a subset of real functions

[jimmybonkers]

let C0 be the set of continuous functions f : R -> R. For n >= 1, let Cn denote theset of functions f : R -> R such that f is differentiable and such that f' is contained in C(n-1). (Therefore Cn is the set of functions whose derivatives f',f'',f''',...,f^(n) up to the nth order exist and are continuous.) Prove by induction that Cn is a subspace of V where V is the set of all functions f : R -> R.

There are three properties that Cn must satisfy to be a subspace,
1.) it must contain the zero vector of V
2.) It must be closed under vector addition
3.) it must be closed under scalar multiplication

I am not sure which of these properties i must perform induction on (obviously not 1.) ) or should it be both 2.) and 3.)..?
I would greatly appreciate it if someone could give me a hint for what the inductive step should be..?

cheers,

James

 

[disregardthat]

Obviosuly Cn is a subset of V, so you just need to prove that it's a vector space by showing that the conditions for a vector space are satisfied. You don't need to do any induction.

 

[jimmybonkers]

I am specifically asked to prove it by induction

 

[Citan Uzuki]

You'll need induction to show that the (f+g)^(n) = f^(n) + g^(n) and that (cf)^(n) = cf^(n) (which is what you need to prove to show that C^n is closed under addition and scalar multiplication). The inductive step would simply be using the linearity of derivation.

 

[Fredrik]

There are three properties that Cn must satisfy to be a subspace,
1.) it must contain the zero vector of V
2.) It must be closed under vector addition
3.) it must be closed under scalar multiplication

I am not sure which of these properties i must perform induction on (obviously not 1.) ) or should it be both 2.) and 3.)..?
I would greatly appreciate it if someone could give me a hint for what the inductive step should be..?

3 implies 1. And 2 and 3 can be combined into "closed under linear combinations".

$$(af+bg)^{(m+1)}=((af+bg)^{(m)})' =\dots$$