# inf sup problem need help

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?

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Jul 22 at 7:25
• The only thing I found is that this infimum is at most $1/e$. This can be shown considering $f(x)=(1-x)^n$ and sending $n \to \infty$. – Crostul Jul 22 at 8:45
• What operation is meant by $*$? Multiplication or convolution? – daw Jul 22 at 9:34

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 $$\epsilon, the $$\sup_x xf(x)$$ is unchanged, while $$\int_0^1 f_\epsilon$$ will converge to $$\int_0^1 f$$ for $$\epsilon\to0$$.