Let $S$={$f_1,f_2,...,f_k$}$\subset V(F)$={$f\vert f:\mathbb R\rightarrow \mathbb R$}.Let $D_1,D_2$ be two subsets of $\mathbb R$ such that $D_1\subset D_2$.I need to know if the following results are true in general
$1.$ if $S$ is linearly independent on $D_1,$ then $S$ is linearly independent on $D_2$.
$2.$ if $S$ is linearly dependent on $D_2,$ then $S$ is linearly independent on $D_1$.
$3.$ if $S$ is linearly dependent on $D_2,$ then $S$ is linearly dependent on $D_1$.
Also,give examples which explains both of the above facts
If $S$ be linearly independent on $D_1$ means that $\{\left.f_1\right|_{D_1}, \ldots, \left.f_n\right|_{D_1}\}$ is l.i. under $\{f;$ $ f: D_1\rightarrow \mathbb{R}\}$, then
1) It is true because $S$ is linearly independent on $D_1$, then
\begin{eqnarray}\alpha_1 f_1(x) +\ldots +\alpha_n f_n(x) = 0, \quad\forall x \in D_1 \quad \Leftrightarrow \alpha_1=\dots=\alpha_n = 0.\quad(I) \end{eqnarray}
Now, suppose that there are $\beta_1, \ldots, \beta_n$ $\in$ $\mathbb{R}$ such that
$$\beta_1 f_1 (x) + \dots + \beta_nf_n(x) =0, \quad\forall x \in D_2$$ using that $D_1$ $\subset$ $D_2$ $$\beta_1 f_1 (x) + \dots + \beta_nf_n(x) =0, \quad\forall x \in D_1.$$ Finally using (I) we conclude that $\beta_1 = ...=\beta_n = 0$, then $S$ is linearly independent on $D_2$.
2) It is false, consider $D_1 = \{0\}$ and $D_2 =\mathbb{R}$, then $S=\{t+1,t+2\}$ is l.i on $D_2 = \mathbb{R}$, but if $f(t) = t+1$ and $g = t +2$, then $2\cdot\left. f\right|_{\{0\}} = \left.g\right|_{\{0\}} $, implying that $S$ is l.d. on $D_1$
3) It is true because if $S$ is linearly dependent on $D_2$, then there are $\alpha_1, ...,\alpha_n$ $\in$ $\mathbb{R}$, not all zero, such that
\begin{eqnarray}\alpha_1 f_1(x) +\ldots +\alpha_n f_n(x) = 0, \quad\forall x \in D_2.\quad(II) \end{eqnarray} using that $D_1 \subset D_2$ and (II), then \begin{eqnarray}\alpha_1 f_1(x) +\ldots +\alpha_n f_n(x) = 0, \quad\forall x \in D_1. \end{eqnarray} Once $\alpha_1,...,\alpha_n$ are not all zeros, we have that $S$ is l.d on $D_1$ as well.