Proof of an Integral inequality Let $f\in C^0(\mathbb R_+,\mathbb R)$ and $a\in\mathbb R_+$, $f^*(x)=\dfrac1x\displaystyle\int_0^xf(t)\,dt$ when $x>0$, and $f^*(0)=f(0)$.
Show that 
$$
\int_0^a(f^*)^2(t)\,dt\le4\int_0^af^2(t)\,dt$$
I tried integration by part without success, and Cauchy-Schwarz is not helping here.
Thanks for your help.

Assume without loss of generality that $f\geqslant0$. Writing $(f^*)^2(t)$ as
$$
(f^*)^2(t)=\frac1{t^2}\int_0^tf(y)\left(\int_0^tf(x)\mathrm dx\right)\mathrm dy=\frac2{t^2}\int_0^tf(y)\left(\int_0^yf(x)\mathrm dx\right)\mathrm dy,
$$
and using Fubini, one sees that
$$
A=\int_0^a(f^*)^2(t)\mathrm dt=2\int_0^af(y)\int_0^yf(x)\mathrm dx\int_y^a\frac{\mathrm dt}{t^2}\mathrm dy,
$$
hence
$$
A\leqslant2\int_0^af(y)\frac1y\int_0^yf(x)\mathrm dx\,\mathrm dy=2\int_0^af(y)f^*(y)\mathrm dy.
$$
Cauchy-Schwarz applied to the RHS yields
$$
A^2\leqslant4\left(\int_0^af(y)f^*(y)\mathrm dy\right)^2\leqslant4\int_0^af^2(y)\mathrm dy\cdot\int_0^a(f^*)^2(y)\mathrm dy=4A\int_0^af^2(y)\mathrm dy,
$$
and the result follows.
