smallest c s.t. $\int_{0}^1f(x^\frac{1}{n})\ dx \le c\int_0^1 f(x)\ dx$

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?

• $\int_{0}^1f(x^\frac{1}{n})\ dx = n \int_0^1f(t)t^{n-1}dt \leq n \int_0^1f(t)dt$ so $c \leq n$ – Conrad Mar 7 at 22:07
• @Conrad I don't understand – Gaboru Mar 7 at 22:45
• I made it an answer since it got too long for a comment to explicit the computations – Conrad Mar 7 at 23:08

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