If $H(b) ≤ b$ then $\int_0^k(v−b)h(b)−[1−H(b)]db≤0 $?

Question

$H$ is a CDF over $[0,1]$ and $v$ is a constant

I want to prove that if $H(b)≤b$ for all $b \in [0,1]$ then :

$$\int_0^k(v−b)h(b)−[1−H(b)]db≤0 $$ for all $k∈[0,1]$

I asked a similar question, but this is another question that I am interested in, Can anyone help me to prove this? or does any theorem helps?

Answer 1

$$\int_0^k((v−b)h(b)−(1−H(b))=db≤0$$ iff

$$vH(k)-\int_0^kbh(b)+\int_0^kH(b)db≤\int_0^k 1db=k.$$

Now, using integration by parts, it is

$$\int_0^kbh(b)db=[bH(b)]_0^k-\int_0^kH(b)db=kH(k)-\int_0^kH(b)db.$$ Thus, the inequality is equivalent to

$$(v-k)H(k)\le k.$$ Since $H(k)\le k$ it is enough to show

$$(v-k)k\le k.$$ In other words

$$vk\le k+k^2$$ which holds because $vk\le k.$