Show that $f(0)=f(-1)$ is a subspace

Question

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 $f(0)=f(-1)$.

Answer 1

$$F(\mathbb{R},\mathbb{R})=\left\{\mathcal{f}:\mathbb{R}\rightarrow \mathbb{R} | \mathcal{f}(0) = \mathcal{f}(-1) \right\}$$ We have to show-

1. It is not a empty set or null element is a part of this set.
2. If $\mathcal{f},\mathcal{g} \in F$, then $\left(\mathcal{f}+\mathcal{g}\right) \in F$.
3. 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: $\mathcal{f}(x)=\left|x+\frac{1}{2}\right|$.

Answer 2

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.