Question

Real Analysis : PLEASE please HELP : If g(x +y) = g(x) +g(y) for all x in R (real numbers). And g(x) > 0 for all x. and g(0) = 1 and g(x) >1 for some x in (0,a), a >0 then g is strictly increasing and continous on R.

Answer 1

I meant g(x+y) = g(x)g(y)

Answer 2

What is the question?

Answer 3

http://www.wolframalpha.com/input/?i=If+g%28x+%2By%29+%3D+g%28x%29+%2Bg%28y%29+for+all+x+in+R.And+g%28x%29+%3E+0+for+all+x.+and+g%280%29+%3D+1+and+g%28x%29+%3E1+for+some+x+in+%280%2Ca%29%2C+a+%3E0+then+g+is+strictly+increasing+and+continous+on+R.

Answer 4

Not sure if this helps, but the only function I know of that has the property g(x+y)=g(x)g(y) is exponentiation.

Answer 5

Perhaps, since we know \(g(x+a)=g(x)\cdot g(a)\) if we let \(x=0\), and \(b\in (0, a)\) such that \(g(b)>1\) we have that \[g(0+b)=g(b)=g(0)\cdot g(b)=1\cdot g(b) > 1\]

Answer 6

thanks so much I'm just thinking quick

Answer 7

how do you show continuity?
or that it's inreasing?

Answer 8

increasing

Answer 9

To show increasing, suppose \(g(b) \geq g(c)\) for some \(b<c\). Then, show that this implies a contradiction

Answer 10

To show continuity, you need to show that for some \(\epsilon \in \mathbb{R}_{>0}\) there exists a \(\delta\in\mathbb{R}_{>0}\) such that \[|x-c|<\delta \implies |f(x)-f(c)|<\epsilon\]For all \(x\in\mathbb{R}\)

Answer 11

And for all \(c\in\mathbb{R}\)

Answer 12

do you know how to do the above fore this question in particular? I think i've shown that g must be increasing... I also know that if g is continous at x-0 then it is continous everywhere, i solved that

Answer 13

for

Answer 14

For it to be continuous at 0, use \(c=0\) in the above definition I gave you.

Answer 15

So show that for every \(\epsilon\in\mathbb{R}_{>0}\) there exists a \(\delta\in\mathbb{R}_{>0}\) such that for all \(x\in\mathbb{R}\)\[|x|<\delta \implies |f(x)-1|<\epsilon\]

I wrote my definition a little bit wonky up there.

Answer 16

if g strictly increasing
f(x)-f(c)=f(x-c)
now (x-c)<delta
f(x-c)<f(delta)
ep=f(delta)??????????

Answer 17

you guys are both saying very similar things to what i got in my solution, I'm just a bit woried about how to use the fact that g(x) >1 for all x in (0,a)

Answer 18

and also the domains (-a,0) (0,a) for arbitrary a doesn't include the point 0