I know this question has been asked here and here.
I see the reason why there are three parts because we need to prove additive identity, closed under addition, and closed under scalar multiplication. That is:
- Take $f_0 : [0,1] \to 0$, show that $f_0 + g = g$
- $f, g\in V$, show that $f+g\in V$
- $f\in V$, and $a\in F$, show that $af\in V$
I saw some answer used $\epsilon-\delta$ Definition of a Limit to prove this problem. I don't understand why the proof of continuous function is a subspace of $\mathbb{R}^{[0,1]}$ need to prove the functions are continuous. It seems to me that the focus is not to prove subspace but continuity.
If possible, can you also work out the proof here?
Best Answer
After reading this and this as well as Epsilon-Delta Definition of a Limit, I understand now.
In the question "Prove the set of continuous real-valued functions on the interval [0,1] is a subspace of $\mathbb{R}^{[0,1]}$", if I write $U$ to be the set of continuous real-valued functions, and $u,w \in U$,
Then the question (only close under addition) can be translated to prove that if you add two continuous functions, are you guaranteed to get a continuous function. i.e.$u + w \in U$ where $U$ is a set of continuous functions.
That's why we need to use Epsilon-Delta Definition of a Limit to prove. Similarly in closed under scalar multiplication.