$f:\mathbb{R}^m\to\mathbb{R}$ differentiable such that $f(x/2) = f(x)/2$, show $f$ is linear. $f:\mathbb{R}^m\to\mathbb{R}$ differentiable such that $f(x/2) = f(x)/2$, show $f$ is linear.
I tried to do:
$f(a+b) = f(2(a+b)/2) = f(2(a+b))/2$
but that wouldn't turn into $f(a)+f(b)$
In the same way: $f(a)/2 + f(b)/2 = f((a+b)/2)$ which won't help.
Also, I need to use the differentiability. I know that there must exist $r(v)$ such that
$$f(a+v) = f(a) + grad(f)\cdot v + r(v)$$
where $\lim_{v\to 0}\frac{r(v)}{|v|} = 0$
 A: We actually only need differentiability at $0$ for this to hold.
First, note that $f(0) = 0$, since the functional equation tells us that $f(0) = f(0)/2$. The existence of a derivative at $0$ tells us that
$$\lim_{v \to 0} \frac{|f(v) - \operatorname{grad}(f) \cdot v|}{\|v\|} = 0.$$
In particular, if we choose any $x \in \mathbb{R}^m$, then this limit converges sequentially for the sequence $v_n = x/2^n$, and we have
$$\lim_{n \to \infty} \frac{|f(x/2^n) - \operatorname{grad}(f) \cdot x/2^n|}{\|x/2^n\|} = 0.$$
The functional equation lets us rewrite $f(x/2^n) = f(x)/2^n$, and everything else in the limit is linear and naturally lets us factor the $2^n$ out of it, so we actually have
$$\lim_{n \to \infty} \frac{|f(x)/2^n - (\operatorname{grad}(f) \cdot x)/2^n|}{\|x\|/2^n} = \lim_{n \to \infty} \frac{|f(x) - \operatorname{grad}(f) \cdot x|}{\|x\|} = 0.$$
But now there is no dependence on $n$ in the limit at all, so the limit can only be $0$ if $f(x) = \operatorname{grad}(f) \cdot x$: in other words, if $f(x)$ is linear with the same constant slope that is its derivative at $0$.
A: Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a differentiable function, then f linear $\Leftrightarrow$ $\dfrac{\partial f(a)}{\partial v}=f(v)$
$\Rightarrow$ $\dfrac{\partial f(a)}{\partial v}=\lim_{t \to 0} \dfrac{f(a+tv)-f(a)}{t}=\lim_{t \to 0} \dfrac{f(a)+tf(v)-f(a)}{t}=f(v)$
$\Leftarrow$ $f(v)=\dfrac{\partial f(a)}{\partial v}=\langle \nabla f(a), v\rangle$ thence, $f(a+tv)=\langle \nabla f(a), a+tv\rangle = \langle \nabla f(a), a\rangle+ t\langle \nabla f(a), v\rangle=f(a)+tf(v)$. $\square$
Now, $f(0)=f(0)/2\rightarrow f(0)=0$. We also have to $f(x/2^n)=f(x)/2^n$ (By induction, it's easy). Then
$$\dfrac{\partial f(0)}{\partial v}=\lim_{t \to 0} \dfrac{f(0+tv)-f(0)}{t}=\lim_{t \to 0} \dfrac{f(tv)}{t}$$
suppose $t=\dfrac{1}{2^n}$, If $t \to 0$, then $n \to \infty$
$$\dfrac{\partial f(0)}{\partial v}=\lim_{t \to 0} \dfrac{f(0+tv)-f(0)}{t}=\lim_{t \to 0} \dfrac{f(tv)}{t}=\lim_{n \to \infty} \dfrac{f(\dfrac{1}{2^n}v)}{\dfrac{1}{2^n}}=\lim_{n \to \infty} \dfrac{\dfrac{1}{2^n}f(v)}{\dfrac{1}{2^n}}=f(v)$$
Showing the result. $\blacksquare$
