Range of Certain Function Involving Combinatorial Coefficients Given any fixed $p\in\{2,3,4,\ldots\}$, define the following function
$$
f_p(x)=\sum_{k=0}^p (-1)^{p-k} {p\choose k} (kx-1)_+^{p-1},
$$
where $y_+=\max(0,y)$.
The goal is to show  $f_p(x)\in (0,1)$ when $x\in (\frac{1}{p},1)$.
Comments:

*

*Numerically it seems that the function $f_p(x)$ is always an increasing function over the interval $(1/p,1)$.


*When $x>1/2$, one can simplify the expression of $f_p(x)$ as follows: for a function $Q: \mathbb{Z}\rightarrow \mathbb{R}$, introduce the shift operator $(\theta Q)(k)=Q(k+1)$.
Setting now $Q(k)= (kx-1)^{p-1}$, then
$$
((1-\theta)^p Q) (0)= \sum_{k=0}^p (-1)^{p-k} {p\choose k} Q(k)=0,
$$
since $Q$ is a polynomial of degree less than $p$, which vanishes under the $p$-times differencing operation $(1-\theta)^p$. So for $x\in (1/2,1)$,
$$
f_p(x)=-\sum_{k=0}^1 (-1)^{p-k} {p\choose k} (kx-1)^{p-1}=1-p(1-x)^{p-1}\in (0,1).
$$
 A: 
Lemma: Let $$g_p(x)=\frac1{(p-1)!}\sum_{k=0}^p(-1)^k\binom pk(x-k)_+^{p-1}.$$ Then $g_p(x)$ is the probability density function of $U_1+U_2+\dots+U_p$, where $U_i$ is uniformly distributed on the interval $[0,1]$ for each $i\in \{1,\dots,p\}$.

For a proof, see Density of the sum of $n$ uniform(0,1) distributed random variables. We can express $f_p$ in terms of $g_p$:
\begin{align}
f_p(x)
  &=x^{p-1}\sum_{k=0}^p (-1)^{p-k}\binom{p}k (k-x^{-1})_+^{p-1}
\\&=x^{p-1}\sum_{k=0}^p (-1)^{k}\binom pk (p-k-x^{-1})_+^{p-1}
\\&=(p-1)!x^{p-1}\cdot \frac1{(p-1)!}\sum_{k=0}^p (-1)^{k}\binom pk (p-x^{-1}-k)_+^{p-1}
\\&=(p-1)! x^{p-1}\cdot g_p(p-x^{-1})
\end{align}
Already, we can see that $f_p(x)>0$ for $x\in (\frac1p,1)$, since for these values of $x$, $p-x^{-1}$ is in the support of $g_p$. We just need to show that $f_p(x)< 1$, which follows in two steps:
$$
f_p(x)=(p-1)! x^{p-1}\cdot g_p(p-x^{-1}) 
\stackrel{1}= 
(p-1)! x^{p-1} \cdot g_p(x^{-1})
\stackrel{2}< 1
$$
Explanation:

*

*Here we use the fact $g_p(p-x)=g_p(x)$. This is obvious since $g_p$ is the pdf of a sum of symmetric random variables.


*Here, we use $$g_p(x)< x^{p-1}/(p-1)! \qquad \text{when }x>1.$$ Proof: By definition, $g_p(x)$ is the $p$-fold convolution of the indicator function $\chi_{(0,1)}$. On the other hand, $x^{p-1}/(p-1)!$ is the $p$-fold convolution of $\chi_{(0,\infty)}$. This makes it apparent that $g_p(x)\le x^{p-1}/(p-1)!$ (if you convolve a bigger function, the result is bigger). Furthermore, strict inequality holds when $x>1$ since then $x$ is big enough for the difference between $\chi_{(0,1)}$ and $\chi_{(0,\infty)}$ to matter.
