[proofplan]
We first verify the second moment directly by additive character orthogonality, both as a base case and to identify the normalization of the Weyl sum. For the higher moments we invoke [Hua's Lemma](/theorems/9065), which is the established finite-differencing estimate for integral cutoffs. Finally, we pass from the real parameter $P$ to the integer cutoff $X=\lfloor P\rfloor$, using $X\le P$.
[/proofplan]
[step:Evaluate the second moment by additive character orthogonality]
Let $X:=\lfloor P\rfloor$. Since $P\ge 1$, one has $1\le X\le P$. Define
\begin{align*}
f_X:[0,1]\to\mathbb C,\qquad \alpha\mapsto \sum_{x=1}^{X}e(\alpha x^k).
\end{align*}
Then $f(\alpha;P)=f_X(\alpha)$.
For $j=1$, expanding the square gives
\begin{align*}
\int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)
=
\sum_{1\le x,y\le X}\int_0^1 e(\alpha(x^k-y^k))\,d\mathcal L^1(\alpha).
\end{align*}
For each integer $m$, the additive character orthogonality identity is
\begin{align*}
\int_0^1 e(\alpha m)\,d\mathcal L^1(\alpha)=\mathbb{1}_{\{0\}}(m).
\end{align*}
Equivalently, this integral is $1$ when $m=0$ and $0$ otherwise. Hence only pairs with $x^k=y^k$ contribute. Since $x,y\in\{1,\dots,X\}$ and $k\ge 2$, the equality $x^k=y^k$ implies $x=y$. Therefore
\begin{align*}
\int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)=X\le P.
\end{align*}
Because $P\ge 1$ and $\varepsilon>0$, we also have $P\le P^{1+\varepsilon}$, so this implies the claimed bound for $j=1$.
[guided]
The purpose of the first step is to identify the basic mechanism behind every later moment estimate: integration over $\alpha$ enforces an integer equation. We have declared
\begin{align*}
f_X:[0,1]\to\mathbb C,\qquad \alpha\mapsto \sum_{x=1}^{X}e(\alpha x^k),
\end{align*}
where $X=\lfloor P\rfloor$, so all sums are finite and all termwise expansions are justified.
For the second moment, expand the product:
\begin{align*}
|f_X(\alpha)|^2=f_X(\alpha)\overline{f_X(\alpha)}=\sum_{1\le x,y\le X}e(\alpha x^k)e(-\alpha y^k).
\end{align*}
Thus
\begin{align*}
|f_X(\alpha)|^2=\sum_{1\le x,y\le X}e(\alpha(x^k-y^k)).
\end{align*}
Because this is a finite sum of continuous functions on $[0,1]$, we may integrate term by term with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure):
\begin{align*}
\int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)
=
\sum_{1\le x,y\le X}\int_0^1 e(\alpha(x^k-y^k))\,d\mathcal L^1(\alpha).
\end{align*}
Now we use orthogonality of additive characters on the unit interval. For an integer $m$, the integral of $\alpha\mapsto e(\alpha m)$ over $[0,1]$ is
\begin{align*}
\int_0^1 e(\alpha m)\,d\mathcal L^1(\alpha)=\mathbb{1}_{\{0\}}(m).
\end{align*}
This equals $1$ when $m=0$ and $0$ otherwise. Here $m=x^k-y^k$. Therefore the integral counts exactly the pairs $(x,y)\in\{1,\dots,X\}^2$ satisfying $x^k=y^k$. Since $x$ and $y$ are positive integers and the map $t\mapsto t^k$ is injective on $\{1,\dots,X\}$, this equation is equivalent to $x=y$. There are exactly $X$ such pairs, so
\begin{align*}
\int_0^1 |f_X(\alpha)|^2\,d\mathcal L^1(\alpha)=X\le P.
\end{align*}
This proves the case $j=1$.
[/guided]
[/step]
[step:Invoke Hua's Lemma for the integer cutoff]
We use [citetheorem:9065]. Applied with the same integer $k\ge 2$, the integer cutoff $X\ge 1$, and the chosen integer $j$ satisfying $1\le j\le k$, Hua's Lemma gives, for every $\varepsilon>0$, a constant $C_{j,k,\varepsilon}>0$ such that
\begin{align*}
\int_0^1 \left|\sum_{x=1}^{X}e(\alpha x^k)\right|^{2^j}\,d\mathcal L^1(\alpha)
\le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}.
\end{align*}
This is an external mean-value estimate for the integer cutoff $X$; the remaining step only transfers it to the real parameter $P$ through $X=\lfloor P\rfloor$.
[/step]
[step:Transfer the integer cutoff estimate to $P$]
Fix an integer $j$ with $1\le j\le k$ and fix $\varepsilon>0$. Applying Hua's Lemma from the previous step to the integer $X=\lfloor P\rfloor$ gives
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)=\int_0^1 |f_X(\alpha)|^{2^j}\,d\mathcal L^1(\alpha)\le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}.
\end{align*}
Since $1\le X\le P$, we have
\begin{align*}
X^{2^j-j+\varepsilon}\le P^{2^j-j+\varepsilon}.
\end{align*}
Therefore
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)
\le C_{j,k,\varepsilon}P^{2^j-j+\varepsilon}.
\end{align*}
This is exactly
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)\lesssim_{j,k,\varepsilon}P^{2^j-j+\varepsilon}.
\end{align*}
The proof is complete.
[guided]
The theorem is stated for a real parameter $P\ge 1$, while Hua's Lemma is stated for an integer cutoff. We bridge that mismatch by setting $X=\lfloor P\rfloor$. Since $P\ge 1$, the integer $X$ satisfies $1\le X\le P$, so it is an admissible cutoff for Hua's Lemma.
By the definition of $f_X$ and the statement notation $f(\alpha;P)$, the two functions agree for every $\alpha\in[0,1]$:
\begin{align*}
f(\alpha;P)=f_X(\alpha).
\end{align*}
Hua's Lemma applies with the fixed integers $k$ and $j$, and with the fixed real number $\varepsilon>0$. It provides a constant $C_{j,k,\varepsilon}>0$, depending only on $j$, $k$, and $\varepsilon$, such that
\begin{align*}
\int_0^1 |f_X(\alpha)|^{2^j}\,d\mathcal L^1(\alpha)
\le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}.
\end{align*}
Substituting $f(\alpha;P)=f_X(\alpha)$ gives
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)
\le C_{j,k,\varepsilon}X^{2^j-j+\varepsilon}.
\end{align*}
The exponent $2^j-j+\varepsilon$ is positive because $j\ge 1$ and $\varepsilon>0$. Hence the inequality $X\le P$ implies
\begin{align*}
X^{2^j-j+\varepsilon}\le P^{2^j-j+\varepsilon}.
\end{align*}
Combining the two estimates yields
\begin{align*}
\int_0^1 |f(\alpha;P)|^{2^j}\,d\mathcal L^1(\alpha)
\le C_{j,k,\varepsilon}P^{2^j-j+\varepsilon}.
\end{align*}
This is the asserted bound with implicit constant depending only on $j$, $k$, and $\varepsilon$.
[/guided]
[/step]
