Page 21.2: The Rolle Theorem

Theorem

If \(y\) has at least \(k\in \mathbb{N}\) GZs on \([a, b]\), counting multiplicities, then \(y^{\Delta}\) has at least \(k-1\) GZs on \([a, b]\), counting multiplicities.

Proof

1. Firstly, we will prove that if \(y\) has a GZ of order \(\leq m\) at \(t\), then \(y^{\Delta}\) has a GZ of order \(\leq m-1\) at \(t\).

Since \(y\) has a GZ of order \(\leq m\) at \(t\), then
\[
y^{\Delta^i}(t)=0,\quad i\in \{0, \ldots, m-1\},
\notag\]
or
\[
y^{\Delta^i}(t)=0,\quad i\in \{0, \ldots, m-2\},\quad \text{and}\quad y^{\Delta^{m-1}}(\rho(t)) y^{\Delta^{m-1}}(t)<0.
\notag\]
If \(m=1\), then \(y(t_0\) or \(y(\rho(t))y(t)<0\). If \(y^{\Delta}(t)=0\) or \(y^{\Delta}(\rho(t))y^{\Delta}(t)<0\), then \(m>1\), which is a contradiction. Therefore \(y^{\Delta}\) has no GZ at \(t\). Let \(m\geq 2\). Then
\[
\left(y^{\Delta}\right)^{\Delta^{i-1}}(t)=0,\quad i\in \{0, \ldots, m-2\},
\notag\]
or
\[
\left(y^{\Delta}\right)^{\Delta^i}(t)=0,\quad i\in \{0, \ldots, m-3\},\quad \text{and}\quad \left(y^{\Delta}\right)^{\Delta^{m-2}}(\rho(t))\left(y^{\Delta}\right)^{\Delta^{m-2}}(t)<0.
\notag\]
Thus, \(y^{\Delta}\) has a GZ of order \(\leq m-1\) at \(t\).

2.  Now, we will prove that if \(y\) has a \(GZ\) of order \(\leq m\in \mathbb{N}\) at \(t\) and \(y\) has a GZ of order \(\leq 1\) at \(s\) with \(\sigma^{m-1}(t)<s\), then \(y^{\Delta}\) has at least \(m\) GZs in \([t, s)\) which is equivalent to
   
a)  If \(y(r)=0\) and \(y^{\Delta}\) has no GZ in \([r, s)\), where \(r<\rho(s)\), then \(y\) has no \(GZ\) at \(s\).

 b) If \(y(\rho(r))y(r)<0\) and \(y^{\Delta}\) has no \(GZ\) in \([r, s)\), where \(r<\rho(s)\), then \(y\) has no  GZ at \(s\).
  
If the assumptions of \(a)\) hold, then \(y^{\Delta}(\tau)>0\), \(\tau\in [r, s)\), or \(y^{\Delta}(\tau)<0\), \(\tau\in [r, s)\). Thus,
    \[\begin{eqnarray*}
    y(\rho(s))y(s)&=& \left(\int_r^{\rho(s)}y^{\Delta}(\tau)\Delta \tau\right)\left(\int_r^s y^{\Delta}(\tau)\Delta\tau\right)\\ \\
    &>& 0.
    \end{eqnarray*}\notag\]
    Therefore \(y\) has no GZ at \(s\). If the assumptions of \(b)\) hold, then \(\rho(r)<r\) and
    \[\begin{eqnarray*}
    y(\rho(r))y^{\Delta}(\rho(r))&=& y(\rho(r))\frac{y(r)- y(\rho(r))}{\mu(\rho(r))}\\ \\
    &<& 0.
    \end{eqnarray*}\notag\]
    Since \(y^{\Delta}\) has a constant sign on \([\rho(r), s)\), then
    \[y(\rho(r))y^{\Delta}(\tau)<0,\quad \tau [\rho(r), s).
    \notag\]
    Therefore
    \[\begin{eqnarray*}
    y(\rho(r))y(t)&=& y(\rho(r))\left(y(r)+\int_r^t y^{\Delta}(\tau)\Delta\tau\right)\\ \\
    &<& 0,\quad t\in \{\rho(s), s\}.
    \end{eqnarray*}\notag\]
    Hence,
    \[
    y(\rho(r))y(\rho(s))<0,\quad y(\rho(r))y(s)<0,
    \notag\]
    and
    \[
    (y(\rho(r)))^2 y(\rho(s))y(s)>0.
    \notag\]
    Consequently \(y(\rho(s))y(s)>0\)  and \(y\) has no GZ at \(s\). This completes the proof.