Questions are in bold.
The set of differentiable real-valued functions on (0,3) such that $f'(2)=b$ is a subspace of $(0,3)\to \mathbb R$ if and only if $b=0$ ($(0,3)\to \mathbb R$ denotes the set of functions from $(0,3)$ to $\mathbb{R}$)
So that it forms a subspace we must have the $0$ element $0(x):x\mapsto0$, and this function have a zero derivative, hence $f'(2)=0$, thus $b=0$.
Also why $f'(2)$ and not $f'(x)$? Right since we have a "if and only if" so that $f'(2)=0$ is not a necessary condition for that $f'(x)=0$ for $x\in(0,3)$?
Also how can we regard a function as a vector?
I understand now that a function from a nonempty interval to $\mathbb{R}$ is a vector in $\mathbb{R}^\infty$ since we can write $$f=\{(x_1,f(x_1)),(x_2,f(x_2)),…\}$$ but do we write $(\mathbb{R}^2)^\infty$ or just $\mathbb{R}^\infty$? Or do we have another notation? ($f$ is real valued)
Best Answer
I'm trying to understand. So, let us call $X$ the set of all functions from $(0,3)$ into $\mathbb{R}$. For a given $b \in \mathbb{R}$, we are give the set $$Y = \left\{ f \in X \mid \hbox{$f$ is differentiable and $f'(2)=b$}\right\}.$$ As you noticed,the zero function belongs to $Y$ if and only if $b=0$. But this is not enough: $Y$ is a vector subspace if and only if $kf \in Y$ whenever $k \in \mathbb{R}$ and $f \in Y$. I'll leave you the easy task to check this necessary and sufficient condition by using the elementary rules of differentiation.
Concerning your question, if I can understand it at all, the condition $f'(2)=b$ could be replaced by $f'(x_0)=b$, where $x_0$ is a fixed point in $(0,3)$. The proof remains unaltered.
Finally, here you are not thinking of such hard things. The canonical way to put a vector space structure on functions is to add them pointwise and multiply them by numbers pointwise. Moreover, $(0,3)$ is uncountable, so that $$\mathbb{R}^{(0,3)} \neq \mathbb{R}^\infty.$$ The latter is the usual symbol for sequences of real numbers, i.e. functions defined on $\mathbb{N}$.