Boundary Crossing Lemma

Boundary Crossing Lemma

Let $\lambda \in [0,1]$ and ${\Delta }_{\lambda }[z]={\Lambda }^{(1)}[z]-\lambda {\Lambda }^{)2}[z]$.

If ${\Lambda }_1[z]$ is Schur and ${\Lambda }_1[z]-{\Lambda }_2[z]$ is not Schur (or vice versa), then there exists ${\lambda }^{\ast }\in [0,1]$ such that ${\Delta }_{{\lambda }^{\ast }}[z]$ has a root on the unit circle.

Visual intuition:

Recall:

Lemma

Suppose $\left|c_n\right|>\left|c_0\right|$. Then $\Delta [z]$ is Schur if and only if $R[z]$ is Schur.

Proof:

• Given: $\left|c_n\right|>\left|c_0\right|$
• WTS: $\Delta [z]$ is Schur $⟺$ $R[z]$ is Schur $⟺$ $zR[z]$ is Schur

For $\lambda \in [0,1]$, let ${\Delta }_{\lambda }[z]=\Delta [z]-\lambda \frac{c_0}{c_n}Q[z]$. We then have

$$⟹{\Delta }_0[z]=\Delta [z]\text{ and }{\Delta }_1[z]=\Delta [z]-\frac{c_0}{c_n}Q[z]=zR[z]$$

WTS: ${\Delta }_0[z]$ is Schur $⟹$ ${\Delta }_1[z]$ is Schur.

Assume toward a contradiction that one of ${\Delta }_0[z],{\Delta }_1[z]$ is Schur and the other is not. By the boundary crossing lemma, there exists ${\lambda }^{\ast }\in [0,1]$ such that ${\Delta }_{{\lambda }^{\ast }}[z]$ has (at least) one root on the unit circle, which we call $p$. Then, we must have:

$${\Delta }_{{\lambda }^{\ast }}[p]=0\text{and}\left|p\right|=1$$

Case 1: $p=1$ or $p=-1$.

$$\begin{array}{r}⟹\frac{1}{p}=p\end{array}$$

Aside

Aside:

$$\begin{array}{rl}\Delta [z]& =z^2+\frac{1}{2}z+\frac{1}{4}\\ \Delta \left[\frac{1}{z}\right]& =\frac{1}{z^2}+\frac{1}{2}\frac{1}{z}+\frac{1}{4}\\ z^2\Delta \left[\frac{1}{z}\right]& =1+\frac{1}{2}z+\frac{1}{4}{z}^2=Q[z]\\ ⟹Q[z]& =z^n\Delta \left[\frac{1}{z}\right]\end{array}$$

Thus, we have:

$$\begin{array}{rlr}0& ={\Delta }_{{\lambda }^{\ast }}[p]=\Delta [p]-\frac{{\lambda }^{\ast }{c}_0}{c_n}Q[p]& \\ & =\Delta [p]-{\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\Delta \left[\frac{1}{p}\right]& \left[Q[z]=z^n\Delta \left[\frac{1}{z}\right]\right]\\ & =\Delta [p]-{\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\Delta [p]& \left[\frac{1}{p}=p\right]\\ & =\Delta [p]\left(1-{\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\right)& \end{array}$$

Aside

Aside:

$$\left|{\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\right|=\underset{\le 1}{\underbrace{\left|{\lambda }^{\ast }\right|}}\underset{<1\text{ [given]}}{\underbrace{\frac{\left|c_0\right|}{\left|c_n\right|}}}\underset{|p|=1}{\underbrace{{\left|p\right|}^n}}<1$$

Thus:

$$\begin{array}{rl}& \Delta [p]=0=\Delta \left[\frac{1}{p}\right]\text{ since }p=\frac{1}{p}\\ & {\Delta }_0[p]=\Delta [p]=0\\ & {\Delta }_1[p]=\Delta [p]-\frac{c_0}{c_n}{p}^n\Delta \left[\frac{1}{p}\right]=0\\ & ⟹p\text{ is an unstable root of both }{\Delta }_0[z]\text{ and }{\Delta }_1[z]\\ & ⟹\text{contradicts the assumption that one of }{\Delta }_0[z]\text{ and }{\Delta }_1[z]\text{ is Schur}\end{array}$$

Case 2: $p=e^{j\theta },\theta \ne 0,\theta \ne \pi$. Then, we have a conjugate $\overline{p}=e^{-j\theta }=\frac{1}{ej\theta }=\frac{1}{p}$. ${\Delta }_{{\lambda }^{\ast }}[z]$ has real coefficients, so its roots come in complex conjugate pairs. Thus, $\overline{p}=\frac{1}{p}$ is also a root of ${\Delta }_{{\lambda }^{\ast }}[z]$:

$${\Delta }_{{\lambda }^{\ast }}\left[\frac{1}{p}\right]=0={\Delta }_{{\lambda }^{\ast }}[p]$$

Then:

$$\begin{array}{rl}0& ={\Delta }_{{\lambda }^{\ast }}[p]=\Delta [p]-{\lambda }^{\ast }\frac{c_0}{c_n}Q[p]=\Delta [p]-{\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\Delta \left[\frac{1}{p}\right]\\ 0& ={\Delta }_{{\lambda }^{\ast }}\left[\frac{1}{p}\right]=\Delta \left[\frac{1}{p}\right]-{\lambda }^{\ast }\frac{c_0}{c_n}Q\left[\frac{q}{p}\right]=\Delta \left[\frac{1}{p}\right]-{\lambda }^{\ast }\frac{c_0}{c_n}\frac{1}{p^n}\Delta [p]\end{array}$$

We can re-arrange the first of the two lines above to become:

$$\Delta [p]={\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\Delta \left[\frac{1}{p}\right]$$

Substituting this into the the second line:

$$\begin{array}{rl}0& =\Delta \left[\frac{1}{p}\right]-{\lambda }^{\ast }\frac{c_0}{c_n}\cancel{\frac{1}{p^n}}\left({\lambda }^{\ast }\frac{c_0}{c_n}\cancel{p^n}\Delta \left[\frac{1}{p}\right]\right)\\ & =\Delta \left[\frac{1}{p}\right]-{\left({\lambda }^{\ast }\frac{c_0}{c_n}\right)}^2\Delta \left[\frac{1}{p}\right]\\ & =\Delta \left[\frac{1}{p}\right]\left(1-{\left({\lambda }^{\ast }\frac{c_0}{c_n}\right)}^2\right)\end{array}$$

Note that we have:

$$\begin{array}{r}{\left|{\lambda }^{\ast }\frac{c_0}{c_n}\right|}^2=\underset{\le 1}{\underbrace{{\left|{\lambda }^{\ast }\right|}^2}}\underset{<1\text{ [given]}}{\underbrace{{\left(\frac{\left|c_0\right|}{\left|c_n\right|}\right)}^2}}<1\\ 0=\Delta \left[\frac{1}{p}\right]\underset{\ne 0}{\underbrace{\left(1-{\left({\lambda }^{\ast }\frac{c_0}{c_n}\right)}^2\right)}}\end{array}$$

which then gives:

$$\Delta \left[\frac{1}{p}\right]=0$$

so:

$$\begin{array}{rl}\Delta [p]& ={\lambda }^{\ast }\frac{c_0}{c_n}{p}^n\cancel{\Delta \left[\frac{1}{p}\right]}=0\\ {\Delta }_0[p]& =\Delta [p]=0\\ {\Delta }_1[p]& =\cancel{\Delta [p]}-\frac{c_0}{c_n}{p}^n\cancel{\Delta \left[\frac{1}{p}\right]}=0\end{array}$$

Thus, $p$ is an unstable root of both ${\Delta }_0[z]$ and ${\Delta }_1[z]$. They are both not Schur, which contradicts the assumption that one of ${\Delta }_0[z]$ and ${\Delta }_1[z]$ is Schur.

Since Case 1 and Case 2 cover all possibilities, it is not possible that ${\Delta }_0[z]$ and ${\Delta }_1[z]$ is Schur (or vice versa). Therefore, we must have that ${\Delta }_0[z]$ is Schur if and only if ${\Delta }_1[z]$ is Schur.