Let $f: \mathbb{R}^{m} \longrightarrow \mathbb{R}$ differentiable such that $\displaystyle f\left(\frac{x}{2}\right) = \frac{f(x)}{2}$ for all $x \in \mathbb{R}^{m}$. Prove that $f$ is linear.

I'm trying to solve without using grad, because in my book, grad is only defined after. But I did not get a good idea, any hints?


By induction we obtain $$ f(x) = 2^n f\left( \frac{x}{2^n} \right) $$

Since $f$ is continuous $$ f(0) = f\left( \lim_{n \to \infty} \frac{1}{2^n} \right) = \lim_{n \to \infty} \frac{1}{2^n} f\left( 1 \right) = 0 $$

Since $f$ is differentiable $$ f(x) = f(0) + df(0)(x) + r(x) $$

where $$ \lim_{x \to 0} \frac{r(x)}{\|x\|} = 0 $$

Therefore for all $n \in \mathbb{N}$ \begin{align} f(x) &= 2^n f\left( \frac{x}{2^n} \right)\\ &= 2^n df(0)\left( \frac{x}{2^n} \right) + 2^n r \left( \frac{x}{2^n} \right)\\ &= df(0)(x) + \|x\| \frac{r \left( \frac{x}{2^n} \right)}{ \frac{\|x\|}{2^n} } \end{align}

Taking the limit $n \to \infty$ $$ f(x) = \lim_{n \to \infty} 2^n f\left( \frac{x}{2^n} \right) = df(0)(x) + \lim_{n \to \infty} \left( \|x\| \frac{r \left( \frac{x}{2^n} \right)}{ \frac{\|x\|}{2^n} } \right) = df(0)(x) $$

Hence $f$ is linear.

  • $\begingroup$ +1 Nice proof!! $\endgroup$ – Lucas Corrêa Apr 14 '18 at 18:41
  • $\begingroup$ +1. Very nice. I changed a bit the last equation if you don't mind. $\endgroup$ – Orest Bucicovschi Apr 14 '18 at 19:04
  • $\begingroup$ @orangeskid Every improvement is welcome. Thanks! $\endgroup$ – mucciolo Apr 14 '18 at 19:35

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