<s>[INST] Tridiagonal matrix inner product inequality I want to show that there is a $c>0$ such that
$$
\left<Lx,x\right>\ge c\|x\|^2,
$$
for alle $x\in \ell(\mathbb{Z})$ where 
$$
L= 
 \begin{pmatrix}
\ddots & \ddots &  &  &  \\
  \ddots & 17 & -4 & 0 &  \\
  \ddots & -4 & 17 & -4 & \ddots  \\
    & 0  & -4 & 17  & \ddots \\
   &   & \ddots & \ddots  &\ddots 
 \end{pmatrix},
$$
is a tridiagonal matrix and
$$
x= 
 \begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix}.
$$
I know that the following holds
\begin{align}
\left<Lx,x\right>&=\left<\begin{pmatrix}
\ddots & \ddots &  &  &  \\
  \ddots & 17 & -4 & 0 &  \\
  \ddots & -4 & 17 & -4 & \ddots  \\
    & 0  & -4 & 17  & \ddots \\
   &   & \ddots & \ddots  &\ddots 
 \end{pmatrix}\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix},\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix}\right>\\
&=\left< -4\begin{pmatrix}
\vdots  \\
  x_{2} \\
  x_1  \\
   x_{0}  \\
   \vdots 
 \end{pmatrix} +17\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix} -4\begin{pmatrix}
\vdots  \\
  x_{0} \\
  x_{-1}  \\
   x_{-2}  \\
   \vdots 
 \end{pmatrix},\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix} \right>\\
&=-4\left<\begin{pmatrix}
\vdots  \\
  x_{2} \\
  x_1  \\
   x_{0}  \\
   \vdots 
 \end{pmatrix},\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix} \right> + 17\left<\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix},\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix} \right> -4\left<\begin{pmatrix}
\vdots  \\
  x_{0} \\
  x_{-1}  \\
   x_{-2}  \\
   \vdots 
 \end{pmatrix},\begin{pmatrix}
\vdots  \\
  x_{1} \\
  x_0  \\
   x_{-1}  \\
   \vdots 
 \end{pmatrix} \right>.
\end{align}
Hence,
$$
\left<Lx,x\right>=-4k+17\|x\|^2, 
$$
where
$$
k=\sum_{j\in \mathbb{Z}}{x_j(x_{j+1}+x_{j-1})}.
$$
Obviously $\|x\|^2\ge 0$, but how can I choose $c$ such that the inequality holds? Here I get stuck, any hints?
 [/INST] If $V$ is the bilateral shift, we have $$L=17 I - 4(V+V^*).$$
From $\|V\|=1$, we get that $V+V^*$ is a selfadjoint with $\|V+V^*\|\leq2$. Then, for a unit vector $x$, 
$$
\langle Lx,x\rangle=17-4\langle(V+V^*)x,x\rangle\geq17-4\|V+V^*\|\geq 17-8=9.
$$
In other words, you can take $c=9$. 
</s>