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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)

user210673
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    Any set that satisfies the axioms of a vector space is a vector space. Even though we first learn about vectors in $\Bbb R^{n}$, it turns out that there are lots of other vector spaces, like you've said: a set of functions can be a vector space. – layman Jan 26 '15 at 12:08
  • @user46944 So a function is a vector? – user210673 Jan 26 '15 at 12:14
  • It can be, if it belongs to a set of functions that satisfies the vector space axioms. – layman Jan 26 '15 at 12:28
  • If I understand your question correctly, you want to prove that if $V = { f: (0,3) \to \Bbb R }$, then the set of differentiable functions with $f'(2) = 0$ is a subspace of $V$, right? So $V$ is just the set of functions from $(0,3)$ to $\Bbb R$, right? It's not a set of differentiable functions? – layman Jan 26 '15 at 12:31
  • @Fundamental from (0,3) to IR, not the opposite – user210673 Jan 26 '15 at 12:46
  • @user46944 I want to prove that the set of differentiable functions from $(0,3)$ to $\mathbb{R}$ such that $f'(2)=b$ is a subspace of $\mathbb{R}\leftarrow(0,3)$ if and only if $b=0$. – user210673 Jan 26 '15 at 12:47
  • @user46944 I understand now that a function from an 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 $\mathbb{R}^\infty$? – user210673 Jan 26 '15 at 12:50
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    Please don't edit your title to make it huge, red or bold, or any strange typography. – Najib Idrissi Jan 26 '15 at 14:15
  • Sorry, but my question is ignored :( – user210673 Jan 26 '15 at 14:16
  • You posted it only two hours ago, have some patience... One of my questions is still awaiting an answer almost two years later, for example. – Najib Idrissi Jan 26 '15 at 14:17
  • With the various edits, it's hard to tell exactly what you're asking, at this point. As for why $f'(2)=0$ rather than $f'(x)=0$ for all $x\in (0,3),$ note that the latter set is simply a subset of the former set. The same result holds there. – Cameron Buie Jan 26 '15 at 14:21
  • @CameronBuie I edited the question and I suppressed the unecessary parts. I don't understand, why in particular $2$? Or we can put any number you like as long as it is in $(0,3)$? – user210673 Jan 26 '15 at 14:23
  • @CameronBuie Also what about a function as a vector, do we say that a function is a vector in $\mathbb{R}^\infty$ or $(\mathbb{R}^2)^\infty$? – user210673 Jan 26 '15 at 14:58
  • See also: https://math.stackexchange.com/questions/1849230/subspaces-of-mathbbr0-3, https://math.stackexchange.com/questions/4343177/vector-subspaces-axler-example-1-35 – DilithiumMatrix Dec 27 '21 at 19:11

1 Answers1

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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 $\mathbb{N}$.

Siminore
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    Everything is now clear, $\Huge\textbf{thank you so so much!!!}$ but something I still don't know is how to interpret a function as a vector, since a function can be written as $$f={(x_1,f(x_1)),(x_2,f(x_2)),...}$$ then can we say that a function is inside the space of all sequences of pairs ? how do we write that $(\mathbb{R}^2)^\infty$? – user210673 Jan 26 '15 at 15:18
  • You are just identifying $f$ with its graph: $\Gamma = {(x,f(x)) \mid x \in (0,3)}$... – Siminore Jan 26 '15 at 15:20
  • and that graph $\Gamma$ is inside which larger vector space? $(\mathbb{R}^2)^\infty$? – user210673 Jan 26 '15 at 15:21
  • $\Gamma \subset \mathbb{R}^2$. Sorry, can you write down your definition of $\mathbb{R}^\infty$? – Siminore Jan 26 '15 at 15:31
  • $\mathbb{R}^\infty$ is the set of all sequences of the form $(x_1,x_2,...)$ where $x_n\in\mathbb{R}$ – user210673 Jan 26 '15 at 15:40
  • You simply cannot enumerate all the points of $(0,3)$. The set of all sequences is totally useless here. – Siminore Jan 26 '15 at 18:28