1. Solving a functional equation

$\displaystyle let \ f:R \to R \ be \ continuous \ with \ f(x+y)= f(x)+f(y) \ , \ \forall x,y \in R$.
$\displaystyle let \ a=f(1) . prove \ that \ f (x)= ax , \forall x \in R$

2. Originally Posted by flower3
$\displaystyle let \ f:R \to R \ be \ continuous \ with \ f(x+y)= f(x)+f(y) \ , \ \forall x,y \in R$.
$\displaystyle let \ a=f(1) . prove \ that \ f (x)= ax , \forall x \in R$
What have you tried? What do you know? These are all important questions. There are ways to do this quicker than others, if you have the right knowledge.

3. Ehm, I looked at some of his earlier posts and it seems he usually only
drops a question and doesn't really seem to care about the answer. He doesn't really want to be taken serious I guess.

4. Originally Posted by Dinkydoe
Ehm, I looked at some of his earlier posts and it seems he usually only
drops a question and doesn't really seem to care about the answer. He doesn't really want to be taken seriously I guess.
Yeah, yeah I know. You'll notice that I answered a lot of them.

5. Originally Posted by Drexel28
Yeah, yeah I know. You'll notice that I answered a lot of them.
Hey. So should I answer this or not? What do you say?

6. Originally Posted by bandedkrait
Hey. So should I answer this or not? What do you say?

7. $\displaystyle f(x)=f(x. 1)=xf(1)=xa=ax$ for all $\displaystyle x\in\mathbb R$.

8. Originally Posted by flower3

$\displaystyle f(x)=f(x. 1)=xf(1)=xa=ax$ for all $\displaystyle x\in\mathbb R$.
Try for every $\displaystyle x\in\mathbb{N}$ or something similar. It follows by induction that $\displaystyle f(n)=f\left(\underbrace{1+\cdots+1}_{n\text{ times}}\right)=nf(1)$ but to say it for $\displaystyle \mathbb{R}$ is not so simple.

9. Okay, I'm solving this purely for personal gratification :-/

$\displaystyle f(x+y) = f(x) + f(y)$

Let $\displaystyle y=0$
so $\displaystyle f(x)=f(x)+f(0)$

i.e. $\displaystyle f(0)=0$

Now, differentiate both sides partially w.r.t x,

$\displaystyle f'(x+y) = f'(x)$................(1)
Again differentiate partially w.r.t y,
$\displaystyle f'(x+y) = f'(y)$................(2)

from (1) and (2)

$\displaystyle f'(x)=f'(y)$

since, x and y are arbitrary, $\displaystyle f'(x)=k$ for all x, where k is a constant (assuming total continuity and differentiability for both $\displaystyle f(x)$ and $\displaystyle f'(x)$

since $\displaystyle f'(x)$ is a constant,

$\displaystyle f(x) = \int(kdx)$

$\displaystyle =kx+c$

since $\displaystyle f(0)=0$ $\displaystyle c=0$ , and since $\displaystyle f(1)=a$, $\displaystyle k=a$

Thus $\displaystyle f(x)=ax$ for all $\displaystyle x$

10. Originally Posted by bandedkrait
Okay, I'm solving this purely for personal gratification :-/

$\displaystyle f(x+y) = f(x) + f(y)$

Let $\displaystyle y=0$
so $\displaystyle f(x)=f(x)+f(0)$

i.e. $\displaystyle f(0)=0$

Now, differentiate both sides partially w.r.t x,

$\displaystyle f'(x+y) = f'(x)$................(1)
Again differentiate partially w.r.t y,
$\displaystyle f'(x+y) = f'(y)$................(2)

from (1) and (2)

$\displaystyle f'(x)=f'(y)$

since, x and y are arbitrary, $\displaystyle f'(x)=k$ for all x, where k is a constant (assuming total continuity and differentiability for both $\displaystyle f(x)$ and $\displaystyle f'(x)$

since $\displaystyle f'(x)$ is a constant,

$\displaystyle f(x) = \int(kdx)$

$\displaystyle =kx+c$

since $\displaystyle f(0)=0$ $\displaystyle c=0$ , and since $\displaystyle f(1)=a$, $\displaystyle k=a$

Thus $\displaystyle f(x)=ax$ for all $\displaystyle x$
You of course know this doesn't work because you assume differentibility. In fact, we only need that it is differentiable at zero to proceed like you did since $\displaystyle \lim_{\Delta x\to 0}\frac{f\left(x+\Delta x\right)-f\left(x\right)}{\Delta x}=\lim_{\Delta x\to 0}\frac{f\left(\Delta x\right)}{\Delta x}=f'(0)$ since $\displaystyle f(0)=f(0+0)=2f(0)\implies f(0)=0$. But, this isn't now you do the problem.

11. Originally Posted by bandedkrait
Okay, I'm solving this purely for personal gratification :-/

$\displaystyle f(x+y) = f(x) + f(y)$

Let $\displaystyle y=0$
so $\displaystyle f(x)=f(x)+f(0)$

i.e. $\displaystyle f(0)=0$

Now, differentiate both sides partially w.r.t x,

$\displaystyle f'(x+y) = f'(x)$................(1)
Again differentiate partially w.r.t y,
$\displaystyle f'(x+y) = f'(y)$................(2)

from (1) and (2)

$\displaystyle f'(x)=f'(y)$

since, x and y are arbitrary, $\displaystyle f'(x)=k$ for all x, where k is a constant (assuming total continuity and differentiability for both $\displaystyle f(x)$ and $\displaystyle f'(x)$

since $\displaystyle f'(x)$ is a constant,

$\displaystyle f(x) = \int(kdx)$

$\displaystyle =kx+c$

since $\displaystyle f(0)=0$ $\displaystyle c=0$ , and since $\displaystyle f(1)=a$, $\displaystyle k=a$

Thus $\displaystyle f(x)=ax$ for all $\displaystyle x$
A question: Aren't you assuming function is differentiable?

12. Originally Posted by aman_cc
A question: Aren't you assuming function is differentiable?

Yes. But I don't see how the relation can be proved otherwise. However, it's more appropriate to use the definition of derivative itself, like drexel28 did.

13. Originally Posted by bandedkrait
Yes. But I don't see how the relation can be proved otherwise. However, it's more appropriate to use the definition of derivative itself, like drexel28 did.
Yes - I saw Drexel28's post. I trust the proof will involve some form of mean value theorem for continuous function. Will be happy if shown that it can be done in a different way as well.

14. I think the best approach is to show for all rational numbers:

Since $\displaystyle f(1) =f(\frac{p}{p}) = p\cdot f(\frac{1}{p}) = a \Rightarrow f(\frac{1}{p}) = \frac{a}{p}$.

It simply follows now for all rational numbers $\displaystyle f(\frac{p}{q}) = p\cdot f(\frac{1}{q}) = a\cdot \frac{p}{q}$

Now with an argument using continuity and the fact that every real number can be written as a infinite sum of rational numbers, it can be shown for all $\displaystyle x\in \mathbb{R}$

15. Originally Posted by Dinkydoe
I think the best approach is to show for all rational numbers:

Since $\displaystyle f(1) =f(\frac{p}{p}) = p\cdot f(\frac{1}{p}) = a \Rightarrow f(\frac{1}{p}) = \frac{a}{p}$.

It simply follows now for all rational numbers $\displaystyle f(\frac{p}{q}) = p\cdot f(\frac{1}{q}) = a\cdot \frac{p}{q}$

Now with an argument using continuity and the fact that every real number can be written as a infinite sum of rational numbers, it can be shown for all $\displaystyle x\in \mathbb{R}$
Exactly. This is how you do it. Let $\displaystyle x\in\mathbb{R}$ be arbitrary. Since $\displaystyle \mathbb{Q}$ is dense in $\displaystyle \mathbb{R}$ there exists some sequence $\displaystyle \left\{q_n\right\}_{n\in\mathbb{N}}$ such that $\displaystyle q_n\to x$. And since $\displaystyle f$ is continuous we know that $\displaystyle x_n\to x\implies f(x_n)\to f(x)$. So, we have that $\displaystyle f(x)=\lim\text{ }f(q_n)=\lim\text{ }\left\{a\cdot q_n\right\}=a\lim\text{ }q_n=ax$.