Le f be a function such that f(x+y)=f(x)+f(y) for all real x,y. Suppose lim as x goes to 0=L . Show that there exists a such that f(x)=ax for all x

Question

Le f be a function such that f(x+y)=f(x)+f(y) for all real x,y. Suppose lim as x goes to 0=L  . Show that there exists  a such that f(x)=ax for all x

zarkon · Answer

Is L supposed to be any number??

zarkon · Answer

Wouldn't you want the limit as x goes to 0 to be 0?

anonymous · Answer

yes sorry  L=0

anonymous · Answer

it was the first part of the question to show, I also showed that $$lim_{x \rightarrow c}f(x)=f(c)$$

zarkon · Answer

what is your guess for the value of a?

anonymous · Answer

1 would be my guess

zarkon · Answer

you can you simplify f(2) as 

$$f(2)=f(1+1)=f(1)+f(1)=f(1)\cdot 2$$

what would f(3),f(4),... and so on be?

zarkon · Answer

*you can simplify...

anonymous · Answer

oh ok let me try

anonymous · Answer

ok so I can get $$f(x)=f(x/2)*2 $$ but then how do I jump to f(x)=ax?

zarkon · Answer

and how did you get that?

zarkon · Answer

can you answer my question from above..what is f(3),f(4),...?

anonymous · Answer

ah yes ok I get it. $$ f(3)=f(2)+f(1)=3*f(1) , f(4)=4f(1)....$$  therefore a=f(1) ? It works nicely for the naturals but for all real?

zarkon · Answer

...we will build up to the reals..

zarkon · Answer

can you show that f(-n)=-f(n)

zarkon · Answer

where n is a natural number

anonymous · Answer

should I start by rewriting $$f(-n) =f(n-2n)$$ or would does it follow from the addition property ?

zarkon · Answer

hint:  0=1+(-1)

anonymous · Answer

ok so $$f(-n)=f(-n)+f(n)-f(n) =f(-n+n)-f(n)=f(0)-f(n)=-f(n)$$

zarkon · Answer

ok  that works...I was thinking 

$$0=f(0)=f(n+(-n))=f(n)+f(-n) \text{ so }f(-n)=-f(n)$$

zarkon · Answer

now you can do negative integers  

$$f(-n)=-f(n)=-f(1)n$$

zarkon · Answer

now can you show 

$$f\left(\frac{1}{m}\right)=f(1)\frac{1}{m}$$

zarkon · Answer

where $m$ is a natural number

anonymous · Answer

should I write f(1/m) as a sum of other terms?

zarkon · Answer

$$\begin{array}{lc} 1= & \underbrace{\frac{1}{m}+\frac{1}{m}+\cdots +\frac{1}{m}}\  & m \text{ terms} \ \end{array}$$

anonymous · Answer

ok so$$f(1)=mf(1/m) \rightarrow\frac{1}{m}f(1)=f(1/m)$$

zarkon · Answer

why does $f(1)=mf(1/m)$?

you need to show that

zarkon · Answer

$$f(1)=f\left(\frac{1}{m}+\frac{1}{m}+\cdots+\frac{1}{m}\right)=f\left(\frac{1}{m}\right)m$$

zarkon · Answer

I assume you used my hint

anonymous · Answer

$$f(1)=f(1/m+1/m...1/m) =f(1/m)+f(1/m)+...+f(1/m) \text{   m times}  $$
$$ f(1)=mf(1/m)  \text{ dividing by m } f(1)*\frac{1}{m}=f(1/m) 
$$

zarkon · Answer

ok..

almost done....


show $$f\left(\frac{n}{m}\right)=f(1)\frac{n}{m}$$

where $n,m\in\mathbb{N},m\neq 0$

zarkon · Answer

we could let $n\in\mathbb{Z}$

zarkon · Answer

or $m\in\mathbb{Z}$

anonymous · Answer

os so $$ f(\frac{n}{m})=f(1/m)+f(1/m)+...f(1/m) \text{  n times} $$
$$f(n/m)=f(1/m)n \text{ using the previous result we have} f(1/m)=f(1)\frac{1}{m} $$
$$f(n/m)=f(1)\frac{n}{m}$$

zarkon · Answer

$m$ obviously not zero

zarkon · Answer

ok

zarkon · Answer

now you proved that $$\lim_{x\to c}f(x)=f(c)$$

correct?

anonymous · Answer

yes

zarkon · Answer

ok...let $x\in\mathbb{R}$

let $$\{x_n\}\subset\mathbb{Q}$$

where $$x_n\to x$$

(we can do this since the rational numbers are dense in the reals)

zarkon · Answer

then $$\lim_{n\to\infty}f(x_n)=f(x)$$

zarkon · Answer

but $$f(x_n)=f(1)x_n$$

by our previous results

...

anonymous · Answer

ok so you are taking Xn as a  subest of the rationals , as subsequence that goes to x?  and then since $$ f(x_n)=f(1)x_n $$ and because of the limit of f(Xn)=f(x) it follows that f(x)=f(1)x?

hence a=f(1)

zarkon · Answer

yes

anonymous · Answer

wow thank you so much! thanks for being so patient !

zarkon · Answer

no problem