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No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\... 1 a) Let$Z = \overline{T(Y)}$. Then$Z$is a closed subspace of a Banach space and hence a Banach space. Define$i = T$. Then$i$is an isometry and$\overline{i(Y)} = \overline{T(Y)} = Z$. So$(Z,i)$is a completion of$Y$. b) Z is a closed subspace of a reflexive space, so it is reflexive. 1 When you say "then$f:L^{\infty}(\Omega)\rightarrow \mathbb{C}$is bounded functional", it means$f\in (L^\infty)^*$, right? Then the convergence of$f_{n_k}\in L^1$is going to be also as elements of$(L^\infty)^*$, and the limit may be not in$L^1$. (Note that$L^1$is not closed in$(L^\infty)^*$in the norm/weak topology.) 1 You cannot apply Banach-Alaoglu for$L^1(\Omega)$, since$L^1(\Omega)$is not the dual space of a normed space. Rather you have to embed$L^1(\Omega)$into larger spaces$L^\infty(\Omega)^*$or$C(\bar\Omega)^*$to obtain a weak-star convergent subsequence. To see that for$p=1$the assertion is not true, consider the sequence$f_n(x)=n \chi_{0,1/n}(x)$on$...
Continuing where you left off, you have two things to show: Does the new sequence $f$ belong to $X$? For this, note that $(f_n)$ is Cauchy and hence norm-bounded. Thus, $\exists R>0$ such that $$\sum_{k=0}^{\infty} (k+1)|f_n(k)|^2 \leq R \quad\forall n\in \mathbb{N}$$ For each $m\in \mathbb{N}$, this implies $$\sum_{k=0}^m (k+1)|f_n(k)|^2 \leq R$$ ...