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Statement

Setting (p. 2). A word is a finite string $A=A(0)A(1)\cdots A(a-1)\in\mathbb C^a$ with $a\in\mathbb N$. Definition 1 gives its aperiodic autocorrelation function on $\{0,1,\ldots,a-1\}$,

$$\Phi_A(n)=\frac1a\sum_{k=0}^{a-1-n}A(k)\overline{A(k+n)},$$

and calls $A$ normalized when $\Phi_A(0)=1$; every binary word, with entries $\pm1$, is normalized. Definition 2 defines the merit factor of $A$ as $1/(2M_A)$, where

$$M_A=\sum_{n=1}^{a-1}\lvert\Phi_A(n)\rvert^2 .$$

For a normalized word the paper sets

$$P_A(z)=\frac1{\sqrt a}\sum_{n=0}^{a-1}A(n)z^n .$$

The norms $\lVert\cdot\rVert_p$ are those of $L^p$ of the unit circle $\{\lvert z\rvert=1\}$ with normalized Haar measure (p. 2).

Lemma 0 (p. 2, quoted). "If $A$ is a normalized word then $\lVert P_A\rVert_2=1$ and $2M_A=\lVert P_A\rVert_4^4-1$."

So for a normalized word the merit factor equals $1/(\lVert P_A\rVert_4^4-1)$, and a sequence of normalized words has merit factors tending to infinity exactly when $\lVert P_A\rVert_4\to1$; the paper draws this reading on p. 3. A binary word of length $a\ge2$ has $\lvert\Phi_A(a-1)\rvert=1/a$, so $M_A>0$ and its merit factor is finite.

Source. T. Downarowicz and Y. Lacroix, "Merit factors and Morse sequences," Theoretical Computer Science 209 (1998), no. 1--2, 377--387, doi:10.1016/s0304-3975(98)00121-2: Definitions 1 and 2 and Lemma 0 with its proof on p. 2 of the authors' 10-page preprint identified on the source card.

Read depth. Claims checked: Definitions 1--2 and Lemma 0 were read clause by clause on the printed page. The proof is the short Parseval computation below and was followed.

Proof pointer

Page 2. Expanding $\lvert P_A\rvert^2$ gives a trigonometric polynomial whose coefficients at $z^n$ and $z^{-n}$, $0\le n\le a-1$, are $\overline{\Phi_A(n)}$ and $\Phi_A(n)$ (the print assigns them the other way round, which makes no difference for real words or for the norms); its constant term gives $\lVert P_A\rVert_2^2=\Phi_A(0)=1$. Applying Parseval to $\lvert P_A\rvert^2$ gives $\lVert P_A\rVert_4^4=1+2\sum_{n=1}^{a-1}\lvert\Phi_A(n)\rvert^2$.

Dependencies

None beyond Parseval's identity on the circle.

Bears on

• Problem 1150: for a polynomial $Q$ of degree $n$ with coefficients $\pm1$ and $A$ its coefficient word, $P_A=Q/\sqrt{n+1}$, and the lemma reads $\lVert Q\rVert_4^4/(n+1)^2=1+2M_A$. With $\lVert P_A\rVert_4^4\le\lVert P_A\rVert_\infty^2\lVert P_A\rVert_2^2$, a uniform bound $M_A\ge\delta>0$ over binary words would give $\max_{\lvert z\rvert=1}\lvert Q(z)\rvert\ge\sqrt{(1+2\delta)(n+1)}$, the problem's gap. The lemma itself proves no such bound; it is the identity that links merit factors to the problem's $L^\infty$ question, and the source card works the comparison out.