Help proving a function is continuously differentiable So i have a function $$f:\mathbb{R} \rightarrow \mathbb{R}$$  that's differentiable at x=0 and for $x,y \in \mathbb{R}$ let  this be true:
$$f(x+2y)=2f(x)f(y)$$
So how can i prove that this function is continuously differentiable? 
I thought i could use any of these theorems: Lagrange's and Rolle's, but i don't know which one and how, perhaps should i show at first it's differentable and then that derivative is continous?
 A: Let $x \in \mathbb{R}$. We want to show that $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists. 
We calculate 
$\frac{f(x+h)-f(x)}{h} = \frac{f(x+2h/2)-f(x + 0)}{h} = \frac{2f(x)f(h/2)-f(x)f(0)}{h}= f(x)\frac{f(h/2)-f(0)}{h}=f(x)\frac{f(\tilde{h})-f(0)}{2\tilde{h}} = \frac{f(x)}{2}\frac{f(\tilde{h})-f(0)}{\tilde{h}}$.
So that the limit $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ equals $\frac{f(x)}{2} f'(0)$.
As $f$ is continuous, this also implies that $f'$ is contiuous as we have
$f'(x) = \frac{f(x)}{2} \cdot f'(0).$
A: I think this is only possible if $f(x)$ is constant for all values of $x$.


*

*Take $y=0$ and let $x$ be arbitrary. Then $$f(x+2\cdot 0) = 2f(x)f(0)$$ which means $f(x)=2f(x)f(0)$ for every $x\in\mathbb R$. Now, if $f(x)=0$ for every $x$, we are done. Otherwise, take some $x$ such that $f(x)\neq 0$, and divide the equation by $f(x)$ to get $1=2f(0)$ or $f(0)=\frac12$.

*Take $x=0$ and let $y$ be arbitrary. Then $$f(0+2y)=2f(0)f(y)$$ which means that $f(2y)=2f(0)f(y)$ for every $y\in\mathbb R$. From 1., we know that $2f(0)=1$, so we have that for every $y\in\mathbb R$, you have $f(2y)=f(y)$.
Using the rule $f(y)=f(2y)$, we can show inductively that for every $x\in\mathbb R$ and every $n\in\mathbb N$, we have $$f(x)=f\left(\frac{x}{2^n}\right)$$

This last equality, along with the fact that $f$ is continuous at $0$ (because if it is differentiable, it is also continuous), can be used to prove that $f(x)=f(0)$ for every $x\in\mathbb R$:


*

*Let $x\in\mathbb R$ be arbitrary, and let $\epsilon > 0$. Then, there exists some $\delta$ such that $|f(y)-f(0)|<\epsilon$ if $|y|<\delta$ (continuity at $0$).

*Let $n$ be such that $\overline x = \frac{|x|}{2^n}<\delta$. We then know that $|f(\overline x) - f(0)| < \epsilon$.

*Then, since $f(x)=f(\overline x)$, we have $|f(x)-f(0)|<\epsilon$.

*In the beginning, the choice of $\epsilon$ was arbitrary. This means that $|f(x)-f(0)|<\epsilon$ for every value of $\epsilon$, something only possible if $f(x)=f(0)$


Since $x$ is arbitrary, $f(x)=f(0)$ is true for every $x$.
