Problem 5. ( chap3. p.87, functional analysis, W.Rudin) I had done part a, b, and d. But I cannot breakthrough part c, and part e. I restate entire problem in the following:
For $0<p<\infty$, let $l^p$ be the space of all functions $x$ (real or complex, as the case may be) on the positive integers, such that $\sum\limits_{n=1}^\infty|x(n)|^p<\infty$.
For $1\le p<\infty$, define $\|x\|_p=\{\sum|x(n)|^p\}^\frac1p$, and define $||x||_\infty =sup_n|x(n)|.
a, Assume $1\le p<\infty$. Prove that $\|x\|_p$ and $\|x\|_\infty$ make $l^p$ and $l^\infty$ into Banach spaces. If $p^{-1}+q^{-1}=1$, prove that $(l^p)^*=l^q$, in the following sense: there is a one-to-one correspondence $\Lambda \leftrightarrow y$ between $(l^p)^*$ and $l^q$, given by $\Lambda x==\sum x(n)y(n)$    $(x\in l^p)$.
b, Assume $1<p<\infty$ and prove that $l^p$ contains sequences that converge weakly but not strongly.
c,On the other hand, prove that every weakly convergent sequence in $l^1$ converges strongly, in spite of the fact that the weak topology of $l^1$ is different from its strong topology (which is induced by the norm).
d,If $0<p<1$, prove that $l^p$, metrized by $d(x,y)=\sum\limits_{n=1}^\infty|x(n)-y(n)|^p$, is a locally bounded F-space which is not locally convex but that $(l^p)^*$ nevertheless separates points on $l^p$. ( Thus there are many convex open sets in $l^p$ but not enough to form a base for its topology.) Show that $(l^p)^*=l^\infty$, in the same sense as in a,. Show also that the set of all $x$ with $\sum|x(n)|<1$ is weakly bounded but not originally bounded.
e, For $0<p\le 1$, let $\tau_p$ be the $weak^*$-topology induced on $l^\infty$by $l^p$; see a, and d,. If $0<p<r\le 1$, show that $\tau_p$ and $\tau_r$are different topologies ( is one weaker than the other ?) but that they induce the same topology on each norm-bounded subset of $l^\infty$.
Hope that someone help me out this problem. Thanks.
 A: Only part e) remains to be discussed.
The dual space of $(l^\infty, \tau_p)$ is $l^p$ by duality theory (recall: if vector spaces $E$ and $F$ are paired by a non-degenerate pairing then the dual space of $E$ with respect to the topology $\sigma(E,F)$ is $F$). Since $l^p \subsetneqq l^r$ for $p \lt r$ this shows that the topologies $\tau_p$ and $\tau_r$ on $l^\infty$ are distinct for $0 \lt p \lt r \leq 1$ since they have distinct continuous linear functionals.
On the other hand, $l^p \subsetneqq l^r$, shows that $\tau_p \subsetneqq \tau_r$. Since the unit ball $B$ of $l^\infty$ is compact with respect to $\tau_1$ by Alaoglu's theorem, we know that the unit ball is compact with respect to the topologies $\tau_p$ for all $0 \lt p \leq 1$ [it is Hausdorff since the coordinate functionals separate points and the map $(B,\tau_1) \to (B,\tau_p)$ is continuous and onto; This shows that $(B,\tau_p)$ is homeomorphic to $(B,\tau_r)$ for $0 \lt p \lt r \leq 1$ by the usual criterion that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism.]
Finally, a norm-bounded subset of $l^\infty$ is contained in $\lambda B$ for some $\lambda \geq 1$, and hence it inherits the subspace topology of $(\lambda B, \tau_p)$, which is independent of $p$. We conclude that the topologies $\tau_p$ and $\tau_r$ coincide on all norm-bounded subsets of $l^\infty$.
