Show properties of the linear operator $L((a_n)_n)=\left(\frac{1}{n}*a_n\right)_n$

Let $L\colon \ell^p \rightarrow \ell^p$ such that $L((a_n)_n)= \left(\frac{1}{n}*a_n\right)_n$.

1) Determine $L'$(adjoint operator), $\ker(L)$, $\ker(L')$, $\operatorname{rg}(L)$, $\operatorname{rg}(L')$ as well as $\operatorname{cl}(\operatorname{rg}(L))$ and $\operatorname{cl}(\operatorname{rg}(L'))$.

2) Let $K$ be a bounded linear operator. Using $L$, show that $\operatorname{cl}(\operatorname{rg}(K')) \subseteq \ker(K)^\perp$ but equality does not hold in general.

3) $K$ surjective $\rightarrow$ $K'$ injective, but $\leftarrow$ does not hold i.g.

4) $K'$ surjective $\rightarrow$ $K$ injective, but $\leftarrow$ does not hold i.g.

So far, I know: $\langle Lx,y \rangle = \langle x,L'y\rangle$ and so it follows $L=L'$. The kernel of $L$ is $(0)$. Then, from the lecture, I get $\operatorname{cl}(\operatorname{rg}(L))= \ker(L')_\perp = (here) \ker(L)_\perp = (0)_\perp = \ell^p$. How can I find $\operatorname{rg}(L)$? I need it to solve $2)$, $3)$ and $4)$.

I appreciate any help.

If $\left(b_n\right)_{n\geqslant 1}$ belongs to $\operatorname{rg}(L)$, then there exists $\left(a_n\right)_{n\geqslant 1}\in\ell^p$ such that for all $n\geqslant 1$, $a_n/n=b_n$ hence $\left(nb_n\right)_{n\geqslant 1}$ belongs to $\ell^p$. Conversely, if $a:=\left(nb_n\right)_{n\geqslant 1}$ belongs to $\ell^p$ then $La=\left(b_n\right)_{n\geqslant 1}$. Therefore, $$\operatorname{rg}(L)=\left\{\left(b_n\right)_{n\geqslant 1}\mid \sum_{n\geqslant 1}n^p \left\lvert b_n\right\rvert^p<+\infty \right\}.$$