Check that $T: \ell^2 \to \ell^2$ is not surjective where $T(x_1, x_2, x_3, \ldots) = (x_1 + x_2, x_2 + x_3, x_3 + x_4, \ldots)$

Question

Let $T: \ell^2 \to \ell^2$ the operator defined as $$T(x_1, x_2, x_3, x_4, \ldots) = (x_1 + x_2, x_2 + x_3, x_3 + x_4, x_4+ x_5, \ldots)$$ where $$\ell^2 = \left\{ (x_n)_{n=1}^\infty : \sum_{n=1}^\infty |x_n|^2 < \infty \right\}$$ I need to prove that $T$ is not surjective. I managed to prove that $T$ is injective and that $T = I + S$, where $I$ is the identity operator and $S$ is the right shift operator. I know that $S$ is not surjective, but no idea how to deal with $T$.

Any help will be appreciated.

Answer 1

Hint: Notice that if $(y_n)_{n=1}^\infty$ is in the image of $T$, then $y_n=x_n+x_{n+1}$, for some sequence $(x_n)_{n=1}^\infty\in\ell ^2$, which implies $$\sum_{n=1}^k(-1)^{n+1}y_n=x_1+(-1)^{k+1}x_{k+1}$$In particular, for $k\rightarrow \infty$, the alternating sum $\sum_{n=1}^\infty(-1)^{n+1}y_n$ converges (to $x_1$ in this case). This is a necessary condition for being in the image of $T$.

Can you think of a sequence $(y_n)_{n=1}^\infty\in\ell^2$ such that $\sum_{n=1}^\infty(-1)^{n+1}y_n$ does not converge?

Edit: You can prove that $\sum^k_{n=1}(-1)^{n+1}y_n=x_1+(-1)^ny_{n+1}$ by induction. Starting with $k=1$ we have simply $$y_1=x_1+(-1)^{k+1}x_{k+1}=x_1+x_2$$which is true by assumption. Assuming the formula is proved for $k$, $$\sum^{k+1}_{n=1}(-1)^{n+1}y_n=\sum_{n=1}^k(-1)^{n+1}y_n+(-1)^{k+2}y_{k+2}=\\x_1+\underbrace{(-1)^{k+1}x_{k+1}+(-1)^{k+2}x_{k+1}}_{=0}+(-1)^{k+2}x_{k+2}=x_1+(-1)^{k+2}x_{k+2}$$

Answer 2

Of course, it is easiest to explicitly exhibit an element in $\ell^2$ that is not in the image of the operator, but as you tagged Functional Analysis, let's actually use it for this for fun. Since you already showed that $T$ is injective and $T=I+S$, it suffices to prove that $-1$ is in the spectrum $\sigma(S)$ of $S$. The spectrum is a closed subset, so it is enough to show that $\{\lambda\in\Bbb C\mid |\lambda|<1\}\subseteq\sigma(S)$. But now it is easy to see that every $z\in \Bbb C$ with $|z|<1$ is in fact an eigenvalue of the left shift $S$, so we are done!