1. ## Functions

Hi everyone,

I am asked to determine all functions f continious in $\mathbb{R}$ such as:

${f(2009)=2009^{2008}\atop (\forall(x;t)\in \mathbb{R}^2) f(x+t)=f(x)+f(t)}$

I don't know how to start?

And thank you anyway.

2. Let $x=t=0\Rightarrow f(0)=2f(0)\Rightarrow f(0)=0$

$f(1)=f\left(\underbrace{\frac{1}{n}+\frac{1}{n}+\l dots+\frac{1}{n}}_{n}\right)=nf\left(\frac{1}{n}\r ight)\Rightarrow f\left(\frac{1}{n}\right)=\frac{1}{n}\cdot f(1)$

$f\left(\frac{m}{n}\right)=\frac{m}{n}\cdot f(1)$

Then $f(x)=f(1)\cdot x, \ \forall x\in\mathbb{Q}$

Let $x\in\mathbb{R}-\mathbb{Q}$. Then exists a sequence $(x_n)_{n\in\mathbb{N}}, \ x_n\in\mathbb{Q}$ such as $\lim_{x\to\infty}x_n=x$

$f(x_n)=f(1)\cdot x_n$

Apply the limit to both sides:

$\lim_{n\to\infty}f(x_n)=f(1)\lim_{x\to\infty}x_n\R ightarrow f(x)=f(1)\cdot x, \ \forall x\in\mathbb{R}$

$f(2009)=2009^{2008}\Rightarrow f(1)\cdot 2009=2009^{2008}\Rightarrow f(1)=2009^{2007}\Rightarrow f(x)=2009^{2007}x$

3. Thank you very much