Characterize all continuous functions such that $\int_0^1 \left(f\left(\sqrt[i]{x}\right)\right)^{k - i}\,dx = {i\over k}$

Question

Let $k \ge 1$ be an odd integer. What are all continuous functions $f: [0, 1] \to \textbf{R}$ such that$$\int_0^1 \left(f\left(\sqrt[i]{x}\right)\right)^{k - i}\,dx = {i\over k}$$for every $i \in \{1, \ldots, k - 1\}$?

Answer 1

The substitution $x = t^i$ leads to$$\int_0^1 (f(t))^{k - i} \cdot t^{i - 1}\,dt = {1\over k},$$for every $i \in \{1, 2, \ldots, k - 1\}$ and also for $i = k$. Therefore,$$\sum_{i = 1}^k (-1)^i \binom{k - 1}{i - 1} \int_0^1 (f(t))^{k - i} t^{i - 1}\,dt = {1\over k} \sum_{i = 1}^k (-1)^i \binom{k - 1}{i - 1} = {1\over k}(1 - 1)^{k - 1} = 0,$$which leads to $$\int_0^1 (f(t) - t)^{k - 1}\,dt = 0.$$From $k - 1 = \text{even} > 0$ and $f - \text{Id}_{[0, 1]}$ continuous we obtain $f = \text{Id}_{[0, 1]}$.