How do you prove that C(0,1) is a vector space?

Question

Say I have this : $C(0,1)=[f:(0,1)\rightarrow \Bbb R| f\space is\space continuous]$. I want to prove that $C(0,1)$ is a vector space.

I know the 9 conditions to prove that the given space is a space vector, but I don't know how to apply them.

Thx for the support

Answer 1

• $\forall_{f,g \in C(0,1)} (f+g)(x) = f(x)+g(x) \in C(0,1)$
• $\forall_{f \in C(0,1)}\forall_{c \in \Bbb{R}} (c \cdot f)(x)=c \cdot f(x) \in C(0,1)$

• Vector Space associativity, commutativity is a consequence of $\Bbb R$ Field associativity, commutativity

• Identity element is $f(x) = 0$

• Inverse element of $f$ is $-f$

• Multiplication compatibility: $$a \cdot(b \cdot f(x))=ab \cdot f(x)$$

• Identity element of scalar multiplication is of course $1$
• Distributivity: $$a \cdot (f+g)(x) = (af + ag)(x)$$ $$(a+b) \cdot f(x) = af(x) +bf(x)$$

$a,b,c$ are scalars

$f,g$ are functions (vectors)

My guess is that your teacher expect you to prove first 2. So you need to use definition of continuous function: $$\forall_{\epsilon > 0} \exists_{\delta>0}\forall_{x_1,x_2 \in D} \quad |x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\epsilon$$ and prove following:

• if $f$ and $g$ meets this definition then $f+g$ does
• if $f$ meets this definition and $c \in \Bbb R$ then $c \cdot f$ does

second one is trivial, with first triangle inequality will help.

The general idea is that when u are given any $\epsilon $ the definition provides you with $\delta_1$ for function $f$ and $\delta_2$ for function g. Now you need to use them to came up with (show that "they exist") with 2 more dletas for function $(f+g)$ and function $(c \cdot f)$