
> 
> Let $\mathscr F$ be the collection of smooth functions $f \colon
>  \mathbb R \to \mathbb R$ such that 
> 
> 
> 1. $f \in C^\infty\_c(\mathbb R)$, with $\text{supp } f \subset [-1,1]$;
> 2. $\int\_0^1 x f(x) dx = \frac{1}{2\pi}$.
> 
> 
> 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?



A one-dimensional integral minimization problem
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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 $\frac1\pi$. 


