When the following functional is bounded? Let $p \geq 1$. I would like to know for what values $p$ the following functional $F$ is bounded in $L_p[0,1]$
$$
F(f) = \int_{0}^{1}f(x^2)dx
$$
My attempt. I can show directly that for $p\in [1,2)$ this functional is unbounded by constructing a sequence of functions
$$
f_n(x) = \frac{1}{c_n}x^{-\frac{1}{2} + \frac{1}{n}},
$$
where $c_n = ||x^{-\frac{1}{2} + \frac{1}{n}}||_{L_p} = \left(\int_0^1\left|x^{-\frac{1}{2} + \frac{1}{n}}\right|^{p}dx\right)^{1/p}$
we have
$$
||f_n|| = 1, \ \ |F(f_n)| \to \infty.
$$
But I really don't know what to do when $p \geq 2$. Also I know that
$$
{L}_{p}^{*} \cong  L_q, 
$$
when
$$
 \frac{1}{p} + \frac{1}{q} =1
$$
but I don't know how to use this here.
 A: For $p > 2$, by change of variables and the Hölder's inequality, one gets
\begin{align}
 \Big|\int_0^1 f(x^2) dx\Big| = \Big|\int_0^1 f(y) \frac{dy}{2\sqrt{y}}\Big| \leqslant \int_0^1 \Big|f(y) \frac{1}{2\sqrt{y}}\Big| dy \leqslant \Big(\int_0^1 \frac{dy}{2y^{\frac{q}{2}}}\Big)^{\frac{1}{q}}\|f\|_p,
\end{align}
where $q$ is the Hölder conjugate of $p$.
If $p > 2$ then $q < 2$. Thus $\int_0^1 y^{-\frac{q}{2}} dy < \infty$. So the function $F$ is bounded for $p > 2$.
For $p = 2$, let $\{f_n\}$ be the sequence of functions defined by
$$\forall x \in (0, 1],\ f_n(x) = \frac{1}{\sqrt{x}(|\log x| + 1)^{1+\frac{1}{n}}}.$$
It is clear that $f_n \in L^2([0, 1])$. In fact, the sequence $\{f_n\}$ is bounded in $L^2([0, 1])$ since
$$ \int_0^1 |f_n(x)|^2\, dx \leqslant \int_0^1 \frac{dx}{x(|\log x| + 1)} < \infty.$$
But
\begin{align}
\forall n \in \mathbb{N}^{\star},\ F(f_n) = \int_0^1 f_n(x^2) dx \gg \int_0^{\frac{1}{e}} \frac{dx}{x(\log \frac{1}{x})^{1+\frac{1}{n}}} \gg \Big[n(\log \tfrac{1}{x})^{-\frac{1}{n}}\Big]_0^{\frac{1}{e}} \gg n.
\end{align}
So $F(f_n)$ goes to the infinity when $n$ tends to $\infty$.
Hence, $F$ is not bounded for $p = 2$.
