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Given natural number $n, k$ consider nondecreasing function $f:\mathbb{N}\cup {0} \to \mathbb{N}\cup {0}$ such that $$ f\left(\sum_{i=1}^n a_i^n\right)=\frac{1}{k} \sum_{i=1}^n f(a_i^n), $$ for arbitrary $a_1,a_2, \ldots,a_n \in \mathbb{N}\cup {0}.$ Find all such functions

For $k=1$ we have usual linear function $f(x)=c \, x.$

For $k>1$ I found trivial solution $f(x)=0.$

Are there other solutions?

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If $k>1$ :

If $f(0) = 0$: By what you say, if $a_i = 0$ for all $i>1$, then we have $$ f(a^n) = \frac{1}{k}f(a^n) \quad\forall a\in \mathbb{N}\cup\{0\} $$ Which means that $f(a^n) = 0$ if $k>1$. If $f$ is non-decreasing, then $0\leq f(a) \leq f(a^n)$, and so $f$ must be identically zero.

If $f(0) \neq 0$: With $a_i=0$ for all $i$, we have $$ f(0) = \frac{n}{k}f(0) $$ so $n=k$. Hence, $$ f(a^n) = \frac{1}{n}f(a^n) + \frac{n-1}{n}f(0) $$ Hence, $f(0) = f(a^n)$ for all $a\in \mathbb{N}\cup\{0\}$. So for any $a\in \mathbb{N}\cup\{0\}$, we have $$ f(0) \leq f(a) \leq f(a^n) = f(0) $$ and so $f$ is constant.

So, in either case, $f$ must be constant.

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  • $\begingroup$ It is true only if $f(0)=0$. $\endgroup$ – Leox Sep 30 '14 at 18:39
  • $\begingroup$ @Leox: Thanks, edited. $\endgroup$ – Prahlad Vaidyanathan Oct 1 '14 at 3:44

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