linear, non-surjective isometry $T:\ell^p(\mathbb{Z})\to \ell^p(\mathbb{Z})$

Question

I need help to find a linear, non-surjective isometry $T:\ell^p(\mathbb{Z})\to \ell^p(\mathbb{Z})$ for $1\le p\le\infty$.

I tried different things, for example if $f:\mathbb{Z}\to\mathbb{Z}, z\mapsto 2z$, then I considered $T(x)=x\circ f$. But $T(x)(z)=x_{f(z)}=x_{2z}$, i.e. $$T((\ldots,x_{-1},x_0,x_1,x_2,\ldots))=(\ldots,x_{-2},x_0,x_2,\ldots)$$and then $T$ need not to be injective. I tried other ideas but I don't have a suitable idea. Could you help me?

Answer 1

$$ (\ldots, x_{-2}, x_{-1}, x_0, x_1, x_2 ,\ldots) \mapsto (\ldots, x_{-2}, x_{-1}, x_0, 0, x_1, x_2 ,\ldots) $$