
MHB Master
#1
January 17th, 2020,
16:29
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:
 for $a,b\in \mathbb{R}$ it holds that $\text{cost}_a+\text{cost}_b=\text{cost}_{a+b}$.
 for $a\in \mathbb{R}$ it holds that $\lambda\text{cost}_a=\text{cost}_{\lambda a}$.
 $(f+g)=(f)+(g)$.
 $f+f=2f$.
 $f+(f)=\text{cost}_0$.
Could you give me a hint how we could these? Aren't all of these trivial?

January 17th, 2020 16:29
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MHB Seeker
#2
January 17th, 2020,
16:49
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:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
And yes, it does look rather trivial.

MHB Master
#3
January 17th, 2020,
18:15
Thread Author
Originally Posted by
Klaas van Aarsen
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:
$$\forall a,b\in\mathbb R : \cost_a+\cost_b=\cost_{a+b}$$
Ahh ok!!
In the same way we could show also the other ones, or not?
$$2. \ \ \ \DeclareMathOperator{\cost}{cost}
\forall a\in\mathbb R,\,\forall x\in\mathbb R : \lambda \cost_a(x)=\lambda a=\cost_{\lambda a}(x)$$
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:
$$f+f=2f$$
$$5. \ \ \ \forall x\in\mathbb R : (f+(f))(x)=(f+(1)f)(x)=f(x)+(1)f(x)=(11)f(x)=0\cdot f(x)=0=\text{cost}_0(x)$$
Therefore:
$$f+(f)=\text{cost}_0$$
Is everything correct? Could we improve something?

MHB Seeker
#4
January 17th, 2020,
18:30
Looks all good to me.

MHB Master
#5
January 17th, 2020,
22:40
Thread Author
Originally Posted by
Klaas van Aarsen
Looks all good to me.
Great!! Thank you!!