Let $f$ be a nonnegative continuously differentiable function on $[0,1]$ with $\int_0^1 f(x) dx = 1$ and $f(1) = 0$.
How small $$C_f := \int_0^1 |(xf(x))'|(1+x) dx$$ can be? In particular, is it possible to choose $f$ such that $C_f < 1$?
For similar problems I usually use the Euler-Lagrange equation, but here I do not know what to do with the absolute value under the integral. Are there any methods for dealing with such problems?
Solution: One can make $C_f$ arbitrarily small, but not reach $0$.
If one wants to make $C_f$ small, one should choose $f$ such that $|(xf(x))'|$ is small for most $x$.
For a constant $a>0$, we choose $$ g_a(x):=\frac{1}{x+a} - \frac{1}{1+a}, \quad b_a := \int_0^1 g_a(x) \mathrm dx, \quad f_a(x):=b_a^{-1} g_a(x). $$ Then $f$ should satisfy the requirements. We also have $b_a\to\infty$ as $a\to 0$.
Using estimates, one can then calculate that $ C_{f_a} \to0 $ holds as $a\to 0$ (I think this is mostly because of $b_a^{-1}\to0$, while the rest can be bounded).
On the other hand, $C_f=0$ is not possible, because then $xf(x)$ would need to be constant. However, this is not possible for a continuously differentiable function on $[0,1]$.