Let $\Bbb{M} $ be the set of decreasing smooth functions in $[0,1]$ for which $f(1)=0.$
Find $$\inf_{f \in \Bbb{M}}\sup_{x \in [0;1]} \frac{x*f(x)}{{\int_{0}^1}f(t)dt} $$
Let now $F(x)=x*f(x)$. I only find out that $ F(0)=0 $ and $F(1)=0$. Only idea is to proof somehow that F is concave and use Rolle's theorem. Can you give some hints?
The inf-sup is zero. Clearly, the quantity in question is non-negative. Let me show that the infimum is indeed zero..
Let $s<1$, define $f(x) = x^{-s}-1$. Then $\int_0^1 f(x) = \frac1{1-s}-1 = \frac s{s-1}$. The function $x f(x)$ has derivative $(1-s)x^{-s}-1$. Hence the product is minimal at $x_s=(1-s)^{\frac1s}$.
Then $$ \frac{\sup_x xf(x) }{\int f dt} = ( (1-s)^{\frac{1-s}s}-(1-s)^{\frac1s}) \cdot \frac{s-1}s = \frac1s ((1-s)^{\frac1s}-(1-s)^{\frac1s+1}) $$ which tends to zero for $s\nearrow 1$.
If you do not like to take this function with pole at zero, you can replace it by some smooth function: Define $f_\epsilon(x) = f(x)$ on $(\epsilon,1]$, $f_\epsilon(x)\le f(x)$ on $(0,\epsilon)$, such that $f_\epsilon$ is smooth and decreasing. For $\epsilon<x_s$, the $\sup_x xf(x)$ is unchanged, while $\int_0^1 f_\epsilon$ will converge to $\int_0^1 f$ for $\epsilon\to0$.