From now on: $L^p := L^p([0,1],m)$, where $m$ is the Lebesgue measure.
Let $f \in L^p$, and let $T(f):= \int_0^1 f(x^2)dx$. Show that $T$ is in $(L^{4})^\ast$ (so it is well defined e bounded) and find $g_T \in L^\frac{4}{3}$ such that $T$ is represented by $g_T$ as $\int f(x)g_T(x)dx$.
Attempt: What I've tried is this, let $t=x^2$ and so $dt=2x dx$, now, we can make the substitution in the integral and let:
$T(f)=\int \frac{f(t)}{2\sqrt{t}}dt$. So to complete proof we have to prove that $\frac{1}{\sqrt{t}}$ is in $L^\frac{4}{3}$, but: $\int t^{-\frac{1}{2} \frac{4}{3}}dt=\int t^{-\frac{2}{3}}dt < \infty $.
Is my proof correct?
As pointed out by Joseph Adams, the fact that $T$ is well-defined and bounded. Moreover, the function $g_T$ should be explicitely mentioned, even if the equality $T(f)=\int \frac{f(t)}{2\sqrt{t}}dt$ makes its expression clear.