Prove that the set of continuous real-valued functions on $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$

Question

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?

Answer 1

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.

Answer 2

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 $af \in C$.

Answer 3

This had me for a loop as well, but since the approach here is correct I had to rethink the strategy it implies. Doing so helped me understand, and so might help others.

The approach presented here does two things:

1. You need to test for a subspace using the definition: additive identity, closed under addition, closed under multiplication.
2. You need to enforce that the thing you are testing is in fact a continuous real valued function.

So in my simple view of things, the solution is like a filter that tests these two points. I think this is what John meant when he said the question could be "translated" to mean that the Epsilon-Delta Limit would satisfy both (subspace and continuous tests).

Feedback welcome, I am new to this and sticking my neck out, but its the fastest way to learn.