Prove that the operator $T:l^{\infty}(\mathbb{R}) \to l^{\infty}(\mathbb{R})$ is compact

Question

Let $T:l^{\infty}(\mathbb{R}) \to l^{\infty}(\mathbb{R})$ be given by
$(T\bar{x})_n := \frac{(\bar{x})_n + (\bar{x})_{n+1}}{2}$.
Prove that $T$ is compact and bounded by $1$.

This is an old exam question I found. I don't understand the "bar" $(\bar{x})_n$ notation. It cannot be complex conjugation since we are in the reals ?

Answer 1

Note that there's no $x$ to apply the bar to. The element you are given is $\bar x $ , so it's just a name.

The operator is not compact, though. If you consider the canonical basis $\{e_n\}$, you have that $Te_n=\frac12\,(e_{n-1}+e_{n})$. So the sequence $\{\frac12\,(e_{n-1})+e_{n}\}$ is in the range of $T$ and it admits no convergent subsequence since any two elements are at distance $1/2$.