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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\...


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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.


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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.)


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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 $...


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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 $$ \sum_{k=0}^m (k+1)|f_n(k)|^2 \leq R $$ ...



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