Minimax problem for optimization $\inf \sup \frac{x f(x)}{\int_0^1 f(t) dt}$ Consider $\bar{C} = C_{d}^0[0,1] $ is the space of continuous decreasing functions, s.a. $f(1) = 0$. We want to determine $\displaystyle \inf_{f \in \bar{C}} \sup_{x\in [0,1]} \frac{x f(x)}{\int_0^1 f(t) dt}$. 
I thought about $\displaystyle xf(x) = xf(x) + 1\cdot f(1) = \int^x_1 f(t)dt - \int_1^x tf'(t)dt$. Actually I even don't know whether derivate exists or not. 
Now we should consider $\dfrac{\int^1_x \left[tf'(t) + f(t)\right] dt}{\int_0^1 f(t)}$. 
Then I thought about Poincare inequality, but fails there. Any hints ? 

Define $f_n(x) = \min(n,{1 \over x}, {n^2\over n-1}(1-x))$.
Note that $f_n(x) = n$ for $x \in [0,{1 \over n}]$, $f_n(x) = {1 \over x}$ for $x \in [{ 1\over n}, 1-{1 \over n}]$ and the graph of $f_n$ is a straight line joining the points $(1-{1 \over n}, {1 \over 1-{1 \over n}}), (1,0)$ for $x \in [{1-{1 \over n}},1]$.
Note that $\sup_{x \in [0,1]} x f_n(x) = 1$ for all $n$.
We have $\int_0^1 f_n(t) dt = 1+ \log(n-1)+ {1 \over 2(n-1)}$.
Hence $\sup_{x \in [0,1]} { x f_n(x) \over \int_0^1 f_n(t) dt} = { 1\over 1+ \log(n-1)+ {1 \over 2(n-1)}}$ and letting $n \to \infty$ we see that the
$\inf$ over $\bar{C}$ is zero.
