What does a "linear space of all continuous functions from R to R" mean?

Question

So far, I have learned that linear spaces are defined by vectors or tensors and the span of linearly independent vectors produces a vector space. But, check this question out:

Let $C(\mathbb{R})$ be the linear space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$. Let $S$ be the set of differentiable functions $u(x)$ that satisfy the differential equation $u' = 2xu + c$ for all real $x$. For which value(s) of the real constant $c$ is this set a linear subspace of $C(\mathbb{R})$?

Here, what does the phrase "linear space of all continuous functions" mean?

Answer 1

Here "linear space" is used synonymous for "vector space", you consider the set of all continuous functions $\Bbb R\to\Bbb R$ with the addition of functions defined by $$(f+g)(x) = f(x) + g(x)$$ and scalar multiples of functions defined by $$(\lambda \cdot f)(x) = \lambda\cdot f(x).$$ Now in this vector space you have to figure out for each $c\in\mathbb R$ if the set of all differentiable $u\colon \Bbb R\to\Bbb R$ satisfying the differential equation $u'=2xu+c$ is a subspace, that is, is it non-empty and closed under addition of functions and taking scalar multiples of functions?

Answer 2

Addition and scalar multiplication are the usual operations on functions. A set of real functions on $\mathbb R$ functions is a vector space of it is closed under addition and scalar multiplication. In this case the zero function belongs to the space only if $c=0$. And if $c=0$ the set of solutions of the DE are closed under addition and scalar multiplication so it is a vector space. [ $u'=2xu,v'=2xv$ imply $(u+v)'=2x(u+v)$ and $(au)'=2x(au)$ ].

Answer 3

Note that you do not need to solve the d.e. to answer this question.

Just suppose that you have two functions $f$ and $g$ which you know are in $S$. So you know that

$f'=2xf+c\\g'=2xg+c$

Then if $S$ is a linear subspace you need $\lambda f$ to be in $S$ for any $\lambda \in \mathbb{R}$. So you need to have

$(\lambda f)' = 2x \lambda f + c$

but $(\lambda f)' = \lambda f' = \lambda(2xf+c)=2x\lambda f + \lambda c$, so this can only work for any $\lambda$ if $c=0$.

Similarly, $f+g$ will only be in $S$ if $c=0$.