A one-dimensional integral minimization problem 
Let $\mathscr F$ be the collection of smooth functions $f \colon
 \mathbb R \to \mathbb R$ such that 
  
  
*
  
*$f \in C^\infty_c(\mathbb R)$, with $\text{supp } f \subset [-1,1]$;
  
*$\int_0^1 x f(x) dx = \frac{1}{2\pi}$. 
  
  
  I would like to compute  $$ \inf_{\mathscr F} \int_0^1 \vert f^\prime
 (x) \vert x^2 \, dx. $$

My (last) hope is that this infimum is $0$ but I do not manage neither to prove it nor to disprove it. 
The point is that the functional I am trying to minimize is invariant under rescaling, in the sense that given $f \in \mathscr F$ the functional computed on $f$ and on $f_\varepsilon(x) = \varepsilon^{-3} f(\frac{x}{\varepsilon})$ (for every $\varepsilon>0$) has the same value. What is this telling?
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This infimum is $\frac1\pi$. Indeed, integrating by parts, for any $f\in\F$ we have
\begin{equation}
 \frac1{2\pi}=I(f):=\int_0^1 x f(x) dx =-J(f)/2,\quad J(f):=\int_0^1 x^2 f'(x) dx,  
\end{equation}
whence 
\begin{equation}
 H(f):=\int_0^1 x^2 |f'(x)| dx\ge|J(f)|=\frac1\pi,  
\end{equation}
so that the infimum in question is $\ge\frac1\pi$.
On the other hand, for any $\ep\in(0,1)$ let 
\begin{equation}
 f_\ep:=c_\ep g_\ep,\quad g_\ep(x):=\frac x\ep\,1_{(0,\ep]}(x)+\frac{1-x}{1-\ep}\,1_{(\ep,1)}(x),
\end{equation}
where $c_\ep$ is such that $I(f_\ep)=\frac1{2\pi}$, so that $c_\ep\to3/\pi$; the convergence everywhere here is for $\ep\downarrow0$. Then $H(f_\ep)\to\frac1\pi$. 
It remains to approximate in $L^1[0,1]$, however closely, both $f_\ep$ and $f'_\ep$ by $f$ and $f'$ for some $f\in\F$; this can be done by using, say, the convolution of $f_\ep$ (or, rather, of a close version of $f_\ep$ with support on $(0,1)$) with a smooth kernel and the multiplication by a constant factor close to $1$. So, the infimum in question is indeed $\frac1\pi$. 
