How to show that the operator $T((v_j)_{j\geqslant1})=(c_jv_j)_{j\geqslant1}$ is compact?

Question

Consider $(l^2,\langle , \rangle)$ and the following operator: $T((v_j)_{j\geqslant1})=(c_jv_j)_{j\geqslant1}$ where $(c_j)_{j\geqslant1}$ satisfies $lim_{j\rightarrow\infty}(c_j)=0$.

I need to show that it is a compact operator, is there a trick or something to help me?

Answer 1

Hint: Approximate $T$ by a sequence of finite-rank operators $T_n$, where $T_n := TP_n$, where $P_n$ is the projection onto the span of the first $n$ unit vectors. Since $\|T-T_n\|\to 0$, the limit is a compact operator. (If you are still stuck, see Conway's A course in functional analysis (second edition), Proposition 4.6 on page 42.)