# Prove the set of continuous real-valued functions on the interval [0,1] is a subspace of R[0,1]

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?

• Do you know that sums of continuous is continuous and multiplication of continuous is continuous? – Will M. Dec 11 '18 at 23:15
• @WillM. Yes I do. But, I am confused that if the sum of continuous is continuous, does it mean its a subspace of R[0,1]? – JOHN Dec 11 '18 at 23:19

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.

First of all, there is nothing about smartness involved here. This is not at all a matter of you being smart or not. However, if you think you are not smart enough, you will not manage to understand. So please change (and improve!) your opinion of yourself.

Now, to the math. Let $$C$$ be the set of continuous functions $$[0,1] \rightarrow \mathbb{R}$$, and $$V$$ be the set of all functions $$[0,1] \rightarrow \mathbb{R}$$.

To show that $$C$$ is a subspace of $$V$$ (if you know that $$V$$ is a vector space over $$\mathbb{R}$$), you need to show that:

1) The neutral element of $$V$$ (that is, the function that is identically zero) is in $$C$$, ie is continuous.

2) $$C$$ is closed under sum, that is, the sum of two continuous functions $$[0,1] \rightarrow \mathbb{R}$$ is again a continuous function $$[0,1] \rightarrow \mathbb{R}$$.

3) If $$f \in C$$ and $$a \in \mathbb{R}$$, then $$af \in C$$.