[proofplan]
We extend the sequence by zero outside its original interval, so shifted copies can be averaged without separate boundary notation. Cauchy-Schwarz bounds the square of the average by the length of the enlarged index interval times the mean square of a short shifted sum. Expanding that square produces diagonal terms, which give the contribution $N^2/H$, and off-diagonal terms, which are regrouped by the shift difference and compared with the displayed correlation sums. The endpoint errors created by the zero extension are bounded using $|z_n|\le 1$ and are absorbed into the same $N^2/H$ term.
[/proofplan]
[step:Extend the sequence by zero and average shifted copies]
Define the finite interval
\begin{align*}
I:=\{n\in\mathbb Z:M<n\le M+N\}.
\end{align*}
Define the zero extension
\begin{align*}
w:\mathbb Z&\to\mathbb C
\end{align*}
by setting $w(n):=z_n$ for $n\in I$ and $w(n):=0$ for $n\notin I$. Let
\begin{align*}
S:=\sum_{M<n\le M+N} z_n=\sum_{n\in\mathbb Z}w(n).
\end{align*}
For each integer $h$ with $0\le h<H$, translation of the summation index gives
\begin{align*}
S=\sum_{n\in\mathbb Z}w(n+h).
\end{align*}
Since $w(n+h)=0$ unless $n+h\in I$, every nonzero term with $0\le h<H$ has
\begin{align*}
M-H+1\le n\le M+N-1.
\end{align*}
Define
\begin{align*}
J:=\{n\in\mathbb Z:M-H<n\le M+N-1\}.
\end{align*}
Then $|J|=N+H-1\le 2N$, because $H\le N$, and
\begin{align*}
S=\frac{1}{H}\sum_{0\le h<H}\sum_{n\in J}w(n+h).
\end{align*}
[/step]
[step:Apply Cauchy-Schwarz to the averaged shifted sum]
By the finite [Cauchy-Schwarz inequality](/theorems/432) applied to the functions $n\mapsto 1$ and
\begin{align*}
n\mapsto \sum_{0\le h<H}w(n+h)
\end{align*}
on the finite set $J$, we obtain
\begin{align*}
|S|^2\le \frac{|J|}{H^2}\sum_{n\in J}\left|\sum_{0\le h<H}w(n+h)\right|^2.
\end{align*}
Using $|J|\le 2N$, this becomes
\begin{align*}
|S|^2\le \frac{2N}{H^2}\sum_{n\in J}\left|\sum_{0\le h<H}w(n+h)\right|^2.
\end{align*}
[guided]
The reason for introducing $J$ is that all shifted sums now have the same outer summation set. This lets us apply Cauchy-Schwarz to one finite sum rather than tracking a different interval for every shift.
We have already shown
\begin{align*}
S=\frac{1}{H}\sum_{n\in J}\sum_{0\le h<H}w(n+h).
\end{align*}
The finite Cauchy-Schwarz inequality says that, for complex numbers $(a_n)_{n\in J}$,
\begin{align*}
\left|\sum_{n\in J}a_n\right|^2\le |J|\sum_{n\in J}|a_n|^2.
\end{align*}
We apply this with
\begin{align*}
a_n:=\sum_{0\le h<H}w(n+h).
\end{align*}
Therefore
\begin{align*}
|S|^2\le \frac{|J|}{H^2}\sum_{n\in J}\left|\sum_{0\le h<H}w(n+h)\right|^2.
\end{align*}
Finally, $J$ contains exactly $N+H-1$ integers, and the hypothesis $H\le N$ gives $N+H-1\le 2N$. Hence
\begin{align*}
|S|^2\le \frac{2N}{H^2}\sum_{n\in J}\left|\sum_{0\le h<H}w(n+h)\right|^2.
\end{align*}
This is the point where the original sum has been converted into second moments of short shifted sums.
[/guided]
[/step]
[step:Expand the square and separate diagonal shift pairs]
For each $n\in J$,
\begin{align*}
\left|\sum_{0\le h<H}w(n+h)\right|^2=\sum_{0\le h_1<H}\sum_{0\le h_2<H}w(n+h_1)\overline{w(n+h_2)}.
\end{align*}
The diagonal terms $h_1=h_2$ contribute
\begin{align*}
\sum_{0\le h<H}\sum_{n\in J}|w(n+h)|^2.
\end{align*}
Since $|w(m)|\le 1$ for every $m\in\mathbb Z$ and $w$ is supported on the $N$-point set $I$, each inner sum is at most $N$. Hence the diagonal contribution is at most $HN$.
For the off-diagonal contribution, using the triangle inequality for complex sums gives
\begin{align*}
\left|\sum_{\substack{0\le h_1, h_2<H, h_1\ne h_2}}\sum_{n\in J}w(n+h_1)\overline{w(n+h_2)}\right|
\le
2\sum_{1\le k<H}\sum_{0\le h<H-k}\left|\sum_{n\in J}w(n+h+k)\overline{w(n+h)}\right|.
\end{align*}
[/step]
[step:Compare each off-diagonal shifted sum with the stated correlation]
Fix integers $k$ and $h$ with $1\le k<H$ and $0\le h<H-k$. Changing variables by setting $m=n+h$ gives
\begin{align*}
\sum_{n\in J}w(n+h+k)\overline{w(n+h)}=\sum_{m\in\mathbb Z}w(m+k)\overline{w(m)}.
\end{align*}
The unrestricted sum is finite because $w$ has finite support. Since $w(m+k)\overline{w(m)}$ is nonzero only when $m\in I$ and $m+k\in I$, we have
\begin{align*}
\sum_{m\in\mathbb Z}w(m+k)\overline{w(m)}=\sum_{M<m\le M+N-k}z_{m+k}\overline{z_m}.
\end{align*}
For each fixed $k$, there are $H-k\le H$ admissible values of $h$. Therefore the off-diagonal contribution is bounded by
\begin{align*}
2H\sum_{1\le k<H}\left|\sum_{M<m\le M+N-k}z_{m+k}\overline{z_m}\right|.
\end{align*}
[/step]
[step:Combine the diagonal and off-diagonal bounds]
Combining the Cauchy-Schwarz estimate with the diagonal and off-diagonal bounds yields
\begin{align*}
|S|^2\le \frac{2N}{H^2}\left(HN+2H\sum_{1\le k<H}\left|\sum_{M<m\le M+N-k}z_{m+k}\overline{z_m}\right|\right).
\end{align*}
Thus
\begin{align*}
|S|^2\le \frac{2N^2}{H}+\frac{4N}{H}\sum_{1\le k<H}\left|\sum_{M<m\le M+N-k}z_{m+k}\overline{z_m}\right|.
\end{align*}
Renaming the dummy shift variable $k$ as $h$ gives
\begin{align*}
\left|\sum_{M<n\le M+N} z_n\right|^2
\le
2\frac{N^2}{H}+4\frac{N}{H}\sum_{1\le h<H}\left|\sum_{M<n\le M+N-h}z_{n+h}\overline{z_n}\right|.
\end{align*}
This is the asserted estimate with an absolute implicit constant.
[/step]
