Show that $f(0)=f(-1)$ is a subspace Let $F(\Bbb R,\Bbb R)$ is a vector-space of functions $f: \Bbb R\to \Bbb R$. The functions  $f(0)=f(-1)$ are an example of subspace?
If it's possible I would like some example of functions such that $f(0)=f(-1)$. 
 A: $$F(\mathbb{R},\mathbb{R})=\left\{\mathcal{f}:\mathbb{R}\rightarrow \mathbb{R} | \mathcal{f}(0) = \mathcal{f}(-1) \right\}$$
We have to show-


*

*It is not a empty set or null element is a part of this set.

*If $\mathcal{f},\mathcal{g} \in F$, then $\left(\mathcal{f}+\mathcal{g}\right) \in F$.

*If $\mathcal{f} \in F$ and $\alpha \in \mathbb{R}$, then $\alpha \mathcal{f} \in F$.


Part-$1$ $\mathcal{f}(x)=0 \ \forall x \in \mathbb{R}$, is the null element. So, $\mathcal{f}(0)=\mathcal{f}(-1)=0$. So, $\mathcal{f} \in F$.
Part-$2$ $$\mathcal{f},\mathcal{g} \in F$$ $$\left(\mathcal{f}+\mathcal{g}\right) (x)=\mathcal{f}(x)+\mathcal{g}(x) $$
$$\left(\mathcal{f}+\mathcal{g}\right) (0)=\mathcal{f}(0)+\mathcal{g}(0)=\mathcal{f}(-1)+\mathcal{g}(-1)=\left(\mathcal{f}+\mathcal{g}\right) (-1)$$ 
So, $\left(\mathcal{f}+\mathcal{g}\right) \in F$.
Part-$3$ $$\mathcal{f} \in F \text{ and } \alpha \in \mathbb{R}$$
$$(\alpha \mathcal{f})(x)=\alpha \mathcal{f}(x)$$
$$(\alpha \mathcal{f})(0)=\alpha \mathcal{f}(0)=\alpha \mathcal{f}(-1)=(\alpha \mathcal{f})(-1)$$
So, $\alpha \mathcal{f}\in F$.
Ex: $\mathcal{f}(x)=\left|x+\frac{1}{2}\right|$.
A: One such function is $f(x) = x(x+1)$. Clearly the zero function is such a function, and any scalar multiple or linear combination of such functions will be such a function. So it is a subspace.
