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#1 h Apr 8, 2008, 11:44 AM • 1 Y

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Prove that if $f: R\to{R}$ is a derivable function with the property $f(x)=f(\frac{x}{2})+\frac{x}{2}f^{'}(x)$ , for every $x\in{R}$ , then f is a polynomial function of degree smaller or equal than 1

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#2 h Mar 23, 2025, 3:51 PM

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Let $f(x) = \sum_{k=0}^{\infty} a_{k}x^{k}$ be a real-valued power series that is differentiable. If $f(x)= f\left(\frac{x}{2}\right) + \frac{x}{2}f'(x)$ holds for all $x \in \mathbb{R}$ , then:

$\sum_{k=0}^{\infty} a_{k}x^{k} = \sum_{k=0}^{\infty} \frac{a_{k}x^{k}}{2^{k}} + \frac{x}{2}\sum_{k=0}^{\infty} a_{k}kx^{k-1} \Rightarrow \sum_{k=0}^{\infty} a_{k}x^{k} = \sum_{k=0}^{\infty} \left(\frac{1}{2^{k}} + \frac{k}{2}\right)a_{k}x^{k} \Rightarrow \frac{1}{2^{k}} + \frac{k}{2} = 1 \Rightarrow k = 0, 1$

and $\frac{1}{2^{k}} + \frac{k}{2} > 1$ for all $k \in \mathbb{N}_{0} \backslash \{0, 1\}$ Hence, $a_{k} = 0$ for all $k \ge 2 \Rightarrow \textcolor{red}{f(x)= a_{0}+a_{1}x}$ is the desired solution to the original functional equation .

This post has been edited 4 times. Last edited by Mathzeus1024, Mar 23, 2025, 4:10 PM

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#3 h Mar 23, 2025, 5:31 PM

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@above: you cannot assume that $f$ is a power series. Not any differentiable function can be represented as a power series...

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