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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.

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The harder part of exercise c is solved here: To see that $\tau_p$ and $\tau_r$ are different use that they have different spaces of continuous linear functionals. It remains to prove that $\tau_p$ and $\tau_r$ induce the same topology on the norm-bounded subsets of $l^\infty$. – Martin Jan 24 '13 at 19:25
@Martin: Thanks you. Could you explain more about part e, for me? – user52523 Jan 24 '13 at 19:35
Instead of copying the entire exercise, it would probably be better to include only the relevant parts for your actual question so that the answerer doesn't have to look for the difficulties. If you could add a few thoughts of your own, that would make a really good question. – Martin Jan 24 '13 at 20:04
up vote 1 down vote accepted

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 $l^\infty$.

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