0

$R=${$f:[0,1]-> \Bbb R$}
R is all the real and continuous functions from [0,1] to $R$,i want to check if there is a additive inverse number relative to +,so is it possible to say that $f(x)+f(x)^-=0\in R$?

Maybe I did not explain myself correctly,I'm trying to prove:
$R=${$f:[0,1]-> \Bbb R$}
(all the continuous and real function )is ring
So the first step is prove $(R,+)$ is abelian group
And that's why I have to prove $f(x)+fx)^-=0$ to get additive inverse number.

Daniel Fischer
  • 206,697
postFix
  • 257

2 Answers2

2

In fact $(R,+)$ is abelian group.

First you should prove

$(i)\forall f,g\in R$, $f+g\in R$

Given $f,g\in R$ then $f+g:[0,1] \to \mathbb{R}$ and since $f,g$ are real then the sum too is.Since $f,g$ are continuous the sum is continuous (is a easy result from calculus ) Then $f+g\in R$

$(ii)\forall f\in R \exists -f\in R:f+(-f)=0$

Given $f\in R$ exists $-f\in R$( because if $f$ is continuous then $-f$ is continuous)where $-f$ is such that $f+(-f)=0$

$(iii)\text{Exists a function $h$ such that for every function $f\in R$, $f+h=f$} $

Since zero function is continous and real then we denote by $0$ the zero function $0=0(x)$ is such that for any $f\in R$ $(f+0)(x)=f(x)+0=f(x)=f$

$(iv)\text{$\forall f,g\in R$, $f+g=g+f$}$

Since the set of functions under the sum is connmutative then any subset too is. Since the set of continuous and real functions is a subset too is connmutative.

$(v) \forall f,g,h\in R, f+(g+h)=(f+g)+h$

It holds since the set of functions under the sum is connmutative and since $R$ is a subset then $R$ inherit that propery

Then $(R,+)$ is abelian group

  • I can tell that because the function is continuous and real is Injective and surjective? – postFix Jan 16 '21 at 13:36
  • @NFLX, you sound very confused. $f$ doesn't have to be injective or surjective. Please try to express your question more clearly. – TonyK Jan 16 '21 at 13:40
  • No. Doesn't exist an implication of continuous function to injective or surjective –  Jan 16 '21 at 13:40
  • I think that was very helpful A function $f:A \to B$ is injective if for each pair $x,y$ in $A$ if $f(x)=f(y)$ then $x=y$.And a surjective if for each $y$ in $B$ there exists $x$ in $A$ such that $f(x)=y$ –  Jan 16 '21 at 13:42
  • 1
    you're welcome I´m glad that it help you –  Jan 16 '21 at 14:52
0

If you freeze $x$, $f(x),g(x),\cdots$ are reals and enjoy the properties of the Abelian group $\mathbb R,+$. These remain $\forall\,x$.

All you have to do is to prove closure, i.e. the sum of two continuous functions is a continuous function.