Page 21.3: The Knesser Theorem

Lemma

Let \(m\in \mathbb{N}\) and \(f\) is \(m\)-times differentiable on \(\mathbb{T}\).

1.  If
\[
\liminf_{t\to\infty} f^{\Delta^m}(t)>0,
\notag\]
then
\[
\lim_{t\to\infty} f^{\Delta^i}(t)=\infty,\quad i\in \{0, \ldots, m-1\}.
\notag\]

2.  If
\[
\limsup_{t\to\infty} f^{\Delta^m}(t)<0,
\notag\]
then
\[
\lim_{t\to\infty} f^{\Delta^i}(t)=-\infty,\quad i\in \{0, \ldots, m-1\}.
\notag\]

Proof

1. Let
\[
0<\lim_{\delta\to 0}\inf_{t\in L_{\delta}(\infty)}f^{\Delta^m}(t)=\alpha.
\notag\]
Suppose that \(\alpha<\infty\).  Then there is an \(\epsilon>0\) such that
\[
f^{\Delta^m}(t)\geq \frac{\alpha}{2},\quad t\in L_{\epsilon}(\infty).
\notag\]
Thus
\[
\int_s^t f^{\Delta^m}(\tau) \Delta \tau\geq \int_s^t \frac{\alpha}{2}\Delta \tau,\quad t, s\in L_{\epsilon}(\infty),\quad s<t,
\notag\]
and
\[
f^{\Delta^{m-1}}(t)-f^{\Delta^{m-1}}(s)\geq \frac{\alpha}{2} (t-s),\quad t, s\in L_{\epsilon}(\infty).
\notag\]
Therefore
\[
\lim_{t\to\infty} f^{\Delta^{m-1}}(t)=\infty.
\notag\]
Now we suppose that \(\alpha=\infty\). Then for any \(c>0\) there exists an \(\epsilon>0\) such that
\[
f^{\Delta^m}(t)\geq \frac{1}{c},\quad t\in L_{\epsilon}(\infty),
\notag\]
and
\[
\int_s^t f^{\Delta^m}(\tau)\Delta\tau \geq \int_s^t \frac{1}{c}\Delta\tau,\quad t, s\in L_{\epsilon}(\infty),\quad s<t,
\notag\]
whereupon
\[
f^{\Delta^{m-1}}(t)-f^{\Delta^{m-1}}(s)\geq \frac{1}{c}(t-s),\quad t, s\in L_{\epsilon}(\infty),\quad s<t.
\notag\]
Therefore
\[
\lim_{t\to\infty} f^{\Delta^{m-1}}(t)=\infty.
\notag\]
By induction, we get
\[
\lim_{t\to\infty} f^{\Delta^i}(t)=\infty,\quad i\in \{0, \ldots, m-1\}.
\notag\]

2.  Let
\[
\beta=\lim_{\delta\to 0} \sup _{t\in L_{\delta}(\infty)}f^{\Delta^m}(t)<0.
\notag\]
Assume that \(\beta>-\infty\). Then there exists an \(\epsilon>0\) such that
\[
f^{\Delta^m}(t)\leq \frac{\beta}{2},\quad t\in L_{\epsilon}(\infty),
\notag\]
and
\[
\int_s^t f^{\Delta^m}(\tau)\Delta \tau\leq \int_s^t \frac{\beta}{2}\Delta \tau,\quad t, s\in L_{\epsilon}(\infty),\quad s<t.
\notag\]
Hence,
\[
f^{\Delta^{m-1}}(t)-f^{\Delta^{m-1}}(s)\leq \frac{\beta}{2}(t-s),\quad t, s\in L_{\epsilon}(\infty),\quad s<t,
\notag\]
and
\[
\lim_{t\to\infty}f^{\Delta^{m-1}}(t)=-\infty.
\notag\]
Assume that \(\beta=-\infty\). Then for any \(c_1<0\) there is an \(\epsilon>0\) such that
\[
f^{\Delta^m}(t)\leq \frac{1}{c_1},\quad t\in L_{\epsilon}(\infty),
\notag\]
and
\[
\int_s^t f^{\Delta^m}(\tau)\Delta \tau\leq \int_s^t \frac{1}{c_1}\Delta \tau,\quad t, s\in L_{\epsilon}(\infty),\quad s<t.
\notag\]
Hence,
\[
f^{\Delta^{m-1}}(t)-f^{\Delta^{m-1}}(s)\leq \frac{1}{c_1}(t-s),\quad t, s\in L_{\epsilon}(\infty),\quad s<t,
\notag\]
and
\[
\lim_{t\to\infty}f^{\Delta^{m-1}}(t)=-\infty.
\notag\]
By induction, we get
\[
\lim_{t\to\infty} f^{\Delta^i}(t)=-\infty,\quad i\in \{0, \ldots, m-1\}.
\notag\]
This completes the proof.

Theorem (The Knesser Theorem)

Let \(n\in \mathbb{N}\) and \(f\) be \(n\) times differentiable on \(\mathbb{T}\). Suppose that there exists an \(\epsilon>0\) such that
\[
f(t)>0,\quad \textrm{sgn}\left(f^{\Delta^n}(t)\right)=s\in \{-1, 1\},\quad t\in L_{\epsilon}(\infty),
\notag\]
and \(f^{\Delta^n}(t)\ne 0\), \(t\in L_{\delta}(\infty)\), for any \(\delta>0\). Then there exists a \(p\in [0, n]\bigcap \mathbb{N}_0\) such that \(n+p\)  is even for \(s=1\) and odd for \(s=-1\) with
\[\begin{eqnarray*}
(-1)^{p+j} f^{\Delta^j}(t)&>& 0,\quad t\in L_{\epsilon}(\infty),\quad j\in [p, n-1]\bigcap \mathbb{N}_0,\\ \\
f^{\Delta^j}(t)&>& 0,\quad t\in L_{\delta_j}(\infty),\quad \delta_j\in (0, \epsilon),\quad j\in [1, p-1]\bigcap \mathbb{N}_0.
\end{eqnarray*}\notag\]

Proof

1.  Let \(s=-1\).  Assume that \(f^{\Delta^{n-1}}(t_1)\leq 0\) for some \(t_1\in L_{\epsilon}(\infty)\). Then there exists a \(t_2\in L_{\epsilon}(\infty)\), \(t_2>t_1\), such that
    \[
    f^{\Delta^{n-1}}(t)\leq f^{\Delta^{n-1}}(t_2)< f^{\Delta^{n-1}}(t_1),\quad t\in L_{\epsilon}(\infty), \quad t>t_2.
    \notag\]
    Hence,
    \[
    \limsup_{t\to\infty} f^{\Delta^{n-1}}(t)<0
    \notag\]
    and by the above lemma, we conclude that
    \[
    \lim_{t\to\infty} f(t)=-\infty.
    \notag\]
    This ia a contradiction. Therefore
    \[
    f^{\Delta^{n-1}}(t)>0,\quad t\in L_{\epsilon}(\infty).
    \notag\]
    Define
    \[\begin{eqnarray*}
    M&=& \bigg\{ m\in \mathbb{N}_0: (-1)^{m+j}f^{\Delta^j}(t)>0,\quad t\in L_{\epsilon}(\infty),\\ \\
    && j\in (m, n-1]\bigcap \mathbb{N}_0,\quad n+m\quad \textrm{odd}\bigg\}.
    \end{eqnarray*}\notag\]
    Then \(n-1\in M\) and there exists \(p=\min M\). If \(p\in \{0, 1\}\) we are done. Assume that \(p\geq 2\). Suppose that \(f^{\Delta^{p-1}}(t)<0\), \(t\in L_{\epsilon}(\infty)\). Suppose that \(f^{\Delta^{p-2}}(t_1)\leq 0\) for some \(t_1\in L_{\epsilon}(\infty)\). Then there exists a \(t_2\in L_{\epsilon}(\infty)\), \(t_2>t_1\), such that
    \[
    f^{\Delta^{p-2}}(t)\leq f^{\Delta^{p-2}}(t_2)< f^{\Delta^{p-2}}(t_1),\quad t\in L_{\epsilon}(\infty),\quad t>t_2.
    \notag\]
    Hence and the above lemma, we obtain that
    \[
    \lim_{t\to\infty} f(t)=-\infty.
    \notag\]
    This is a contradiction. Therefore \(f^{\Delta^{p-2}}(t)>0\), \(t\in L_{\epsilon}(\infty)\). Also,
    we have that
    \[
    (-1)^{p-2+j} f^{\Delta^j}(t)>0,\quad t\in L_{\epsilon}(\infty),\quad j\in [p-2, n-1]\bigcap \mathbb{N}_0,
    \notag\]
    which is a contradiction by the definition of \(p\). Therefore there exists a \(t_3\in L_{\epsilon}(\infty)\) so that
    \[
    f^{\Delta^{p-1}}(t_3)\geq 0.
    \notag\]
    Because \(f^{\Delta^p}(t)>0\), \(t\in L_{\epsilon}(\infty)\), we conclude that
    \[
    f^{\Delta^{p-1}}(t)>0,\quad t\in L_{\epsilon}(\infty),\quad t>t_3.
    \notag\]
    Hence,
    \[
    \liminf_{t\to\infty}f^{\Delta^{p-1}}(t)>0
    \notag\]
    and by the above lemmaLemma, we obtain
    \[
    \lim_{t\to\infty} f^{\Delta^j}(t)=\infty,\quad 0\leq j\leq p-2.
    \notag\]
   

2.  Let \(s=1\). Assume that there exists a \(t_4\in L_{\epsilon}(\infty)\) such that \(f^{\Delta^{n-1}}(t_4)\geq 0\). Then there exists a \(t_5\in [t_4, \infty)\), \(t_5>t_4\), such that
        \[
        f^{\Delta^{n-1}}(t)\geq f^{\Delta^{n-1}}(t_5)> f^{\Delta^{n-1}}(t_4),\quad t\in L_{\epsilon}(\infty),\quad t>t_5.
        \notag\]
        Hence,
        \[
        \liminf_{t\to\infty} f^{\Delta^{n-1}}(t)>0
        \notag\]
        and by the above lemma, we conclude that
        \[
        \liminf_{t\to\infty}f^{\Delta^j}(t)=\infty,\quad j\in \{1, \ldots, n-2\}.
        \notag\]
        So, \(p=n\). This completes the proof.