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Let $F=\{f:[0,1]\to[0,\infty)|f \ \mathrm{continuous}\}$ and $n \ge 2$ (natural). Determine the smallest constant $c$ s.t. $$ \int_0^1 f\left(x^{1/n}\right)\ dx \le c\int_0^1 f(x)\ dx, \quad \forall f\in F $$

I said that $f_p:[0,1]\to[0,1],f_p(x)=x^p$ is in $F$.

So $$\int_{0}^1f \left(x^{p/n}\right)\ dx \le c\int_0^1 x^p\ dx$$ and $$ \frac{n}{n+p} \le \frac{c}{p+1} \quad p \to \infty \implies c \ge n. $$

To find if $c \ge n$ I tried to substitute $x^{1/n}=t$ so $$n \int_0^1f(t)t^{n-1}dt \le c \int_0^1f(t)dt,$$ but I don't think this will help. Can somebody give me some tips, please?

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    $\begingroup$ $\int_{0}^1f(x^\frac{1}{n})\ dx = n \int_0^1f(t)t^{n-1}dt \leq n \int_0^1f(t)dt$ so $c \leq n$ $\endgroup$ – Conrad Mar 7 at 22:07
  • $\begingroup$ @Conrad I don't understand $\endgroup$ – Gaboru Mar 7 at 22:45
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    $\begingroup$ I made it an answer since it got too long for a comment to explicit the computations $\endgroup$ – Conrad Mar 7 at 23:08
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Since $0 \leq t \leq 1, n-1 \geq 1$, we have $0 \leq t^{n-1} \leq 1$; now $f(t) \geq 0$ so multiplying we get the inequality $f(t)t^{n-1} \leq f(t)$ on the whole interval $[0,1]$; integrating we get $ \int_0^1f(t)t^{n-1}dt \leq \int_0^1f(t)dt$, then multiplying by $n$, and using the change of variables that you noted, gives the inequality $\int_{0}^1f(x^\frac{1}{n})\ dx \leq n \int_0^1f(t)dt$, so if we rename now $t$ as $x$ (change variables again trivially $t=x$) we get $\int_{0}^1f(x^\frac{1}{n})\ dx \leq n \int_0^1f(x)dx$ which means that $n$ satisfies the required inequality; since $c$ is by definition the least such number, this implies $c \leq n$ and this together with what you did solves the problem

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  • $\begingroup$ Thank you! It wasn't so hard at all. $\endgroup$ – Gaboru Mar 8 at 5:07
  • $\begingroup$ You almost did it completely, just one extra step needed $\endgroup$ – Conrad Mar 8 at 11:56

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