Show the properties of the maps

[mathmari]

Hey!

Let $a\in \mathbb{R}$. We define the map $\text{cost}_a:\mathbb{R}\rightarrow \mathbb{R}$, $x\mapsto a$. We define also $-f:=(-1)f$ for a map $f:\mathbb{R}\rightarrow \mathbb{R}$.

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a map and $\lambda\in \mathbb{R}$.

Show that:

1. for $a,b\in \mathbb{R}$ it holds that $\text{cost}_a+\text{cost}_b=\text{cost}_{a+b}$.
2. for $a\in \mathbb{R}$ it holds that $\lambda\text{cost}_a=\text{cost}_{\lambda a}$.
3. $-(f+g)=(-f)+(-g)$.
4. $f+f=2f$.
5. $f+(-f)=\text{cost}_0$.

Could you give me a hint how we could these? Aren't all of these trivial? (Wondering)

 

[I like Serena]

Hey mathmari!

I guess the first one can be shown as follows:
$$\DeclareMathOperator{\cost}{cost}
\forall a,b\in\mathbb R,\,\forall x\in\mathbb R : (\cost_a+\cost_b)(x)=\cost_a(x)+\cost_b(x)=a+b=\cost_{a+b}(x)$$
Therefore:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
(Thinking)

And yes, it does look rather trivial. (Tauri)

 

[mathmari]

I guess the first one can be shown as follows:
$$\DeclareMathOperator{\cost}{cost}
\forall a,b\in\mathbb R,\,\forall x\in\mathbb R : (\cost_a+\cost_b)(x)=\cost_a(x)+\cost_b(x)=a+b=\cost_{a+b}(x)$$
Therefore:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
(Thinking)

Ahh ok! (Malthe)

In the same way we could show also the other ones, or not? (Wondering)

$$2. \ \ \ \DeclareMathOperator{\cost}{cost}
\forall a\in\mathbb R,\,\forall x\in\mathbb R : \lambda \cost_a(x)=\lambda a=\cost_{\lambda a}(x)$$
Therefore:
$$\forall a\in\mathbb R : \lambda\cost_a=\cost_{\lambda a}$$

$$3. \ \ \ \forall x\in\mathbb R : -(f+g)(x)=(-1)(f+g)(x)=(-1)(f(x)+g(x))=(-1)f(x)+(-1)g(x)=(-f)(x)+(-g)(x)$$
Therefore:
$$-(f+g)=(-f)+(-g)$$

$$4. \ \ \ \forall x\in\mathbb R : (f+f)(x)=f(x)+f(x)=2f(x)$$
Therefore:
$$f+f=2f$$

$$5. \ \ \ \forall x\in\mathbb R : (f+(-f))(x)=(f+(-1)f)(x)=f(x)+(-1)f(x)=(1-1)f(x)=0\cdot f(x)=0=\text{cost}_0(x)$$
Therefore:
$$f+(-f)=\text{cost}_0$$ Is everything correct? Could we improve something? (Wondering)

 

[I like Serena]

Looks all good to me. (Nod)

 

[mathmari]

Looks all good to me. (Nod)

Great! Thank you! (Yes)