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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?

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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 $c=9$.

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You can also finish your proof by noting that $k \le 2 \, \|x\|^2$ (by applying Hölder's inequality). Hence, $$\langle L \, x , x \rangle \ge -4 \, k + 17 \, \|x\|^2 \ge 9 \, \|x\|^2.$$

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