Is set of all contiuous functions subspace? This is one of the problems from the book: Hoffman and Kunze, chapter: Vector Spaces
Let $V$ be the (real) vector space of all functions $f$ from $\mathbb{R}$ into $\mathbb{R}$. Is the set of all $f$ which are continuous, subspace of $V$?
I am not sure how to proceed with the solution.
 A: You should apply the subspace test. Suppose $V$ is the vector space of all functions from $\mathbb{R}$ to $\mathbb{R}$, and $W$ is the subset of all continuous functions in $V$. To see if $W$ is a subspace of $V$ it's enough to check that $W$ is non-empty, and that it's closed under vector addition and scalar multiplication. 
It's clear that $W$ is non-empty, since, for example, it contains the constant zero function. 
Showing that $W$ is closed under scalar multiplication means checking that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and $c\in \mathbb{R}$ is a real number, then the function $c.f : x \mapsto cf(x)$ is also continuous. Showing that $W$ is closed under vector addition means checking that if $f$ and $g$ are continuous functions (belonging to $W$), then so is $f+g: x \mapsto f(x)+g(x)$.
The essential facts needed to prove these two properties are the following: the addition and multiplication functions $+: \mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$ and $.: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ are both continuous. If $f$ and $g$ are two continuous functions from $\mathbb{R}$ to $\mathbb{R}$, then the composition function $f\circ g: x \mapsto f(g(x))$ is also continuous.
A: *

*Note that if $f$ is a continuous function then $\alpha \cdot f$ is also a continuous function.

A: Yes, if $f$ is continuous so for any reel number $\alpha$ we have in a point $x\in\mathbb{R}$ 
$$
\forall \epsilon'>0 \qquad \exists r_{\epsilon'}>0 \qquad |y-x|\leq r_{\epsilon'} \Rightarrow |f(x)-f(y)|\leq \epsilon'  \qquad *
$$ 
So we put $g=\alpha.f$ and we proof that $g$ is continuous in $x$ so for $\epsilon>0$ we put $\epsilon'=\epsilon/\alpha$ then in $*$ we find :
$$
\exists r_{\epsilon'}>0 \qquad |y-x|\leq r_{\epsilon'} \Rightarrow |f(x)-f(y)|\leq \epsilon/\alpha \qquad  
$$ 
that's mean :
$$
\exists r_{\epsilon'}>0 \qquad |y-x|\leq r_{\epsilon'} \Rightarrow |g(x)-g(y)|\leq \epsilon
$$ 
So $g$ is continuous in $x$ eq $\alpha.f$ is continuous.
The sum is more complicated, so if we take $f$ and $g$ two continuous functions we put $h=f+g$ so we have in a point $x\in \mathbb{R}$ :
$$
\forall \epsilon'>0 \qquad \exists r_{\epsilon'}>0 \qquad |y-x|\leq r_{\epsilon'} \Rightarrow |f(x)-f(y)|\leq \epsilon'  \qquad **
$$ 
$$
\forall \epsilon''>0 \qquad \exists r'_{\epsilon''}>0 \qquad |y-x|\leq r'_{\epsilon''} \Rightarrow |g(x)-g(y)|\leq \epsilon'  \qquad ***
$$ 
so for $\epsilon>0$ we put $\epsilon'=\epsilon''=\epsilon/2$ then in $**$ and $***$ we find :
$$
\exists r_{\epsilon'}>0 \qquad |y-x|\leq r_{\epsilon'} \Rightarrow |f(x)-f(y)|\leq \epsilon/2  \qquad **'
$$ 
$$
\exists r'_{\epsilon''}>0 \qquad |y-x|\leq r'_{\epsilon''} \Rightarrow |g(x)-g(y)|\leq \epsilon/2  \qquad ***'
$$
we put $r_\epsilon=\min\{r_{\epsilon'};r_{\epsilon''} \}$ so if $|y-x|\leq r_\epsilon$ we have :
$$
|h(x)-h(y)|=|f(x)-f(y)-(g(x)-g(y))|\leq |f(x)-f(y)|+|g(x)-g(y)|\leq \epsilon/2+\epsilon/2=\epsilon
$$
So we have proof that :
$$
\forall \epsilon>0 \qquad \exists r_{\epsilon}>0 \qquad |y-x|\leq r_{\epsilon} \Rightarrow |h(x)-h(y)|\leq \epsilon  
$$ 
and so $f+g$ is continuous.
Conclusion : The set of continuous functions is a vector space over the reel field.
