Topological dual of a topological vector space I am solving a problem from Rudin's Functional Analysis book, where I have to show that the topological dual of $\ell^p$, $0<p<1$ is isometrically isomorphic to $\ell^{\infty}$. The metrics on $\ell^p$ and $\ell^{\infty}$. The metric used on $\ell^p$ is defined as $d(x,y)=\sum\limits_{n=1}^{\infty}|x(n)-y(n)|^p$ for all $x,y\in \ell^p$.
I have defined $F:\ell^{\infty}\to (\ell^{p})^*$ by $F(y)=f_y$ for all $y\in \ell^{\infty}$, where $f_y(x)=\sum\limits_{n=1}^{\infty}x(n)y(n)$. Clearly, $F$ is linear and it can be shown that $F$ is onto also. But how to show that $F$ is an isometry? I got stuck here. Any help is appreciated.
 A: I believe you meant $F\colon \ell ^{\infty}\rightarrow (\ell ^p)^*$, not just $\ell ^p$.  In any case, this is indeed an isomorphism in the category of topological vector spaces when $(\ell ^p)^*$ is equipped with the strong polar topology (the topology of uniform convergence on weakly-bounded subsets of $\ell ^p)$.
First note that a subset of $\ell ^p$ is weakly-bounded iff it is bounded in the $\ell ^1$ norm.  To see this, consider the following.  We already know that $(\ell ^p)^*\cong _{\mathbf{Vect}}\ell ^{\infty}$, and so $B\subseteq \ell ^p$ is weakly-bounded iff $y(B)\subseteq \mathbb{C}$ is bounded for all $y\in \ell ^{\infty}$.  As $\ell ^p\subseteq \ell ^1$, we may regard $B$ as a subset of $\ell ^1$, and as we also know that $(\ell ^1)^*\cong \ell ^{\infty}$, we likewise have that $B\subseteq \ell ^1$ is weakly-bounded, and hence bounded by Mackey's Theorem.  Thus, the strong polar topology on $(\ell ^p)^*$ is the topology of uniform convergence on $\ell ^1$-bounded subsets of $\ell ^p$.
You seem to already know how to check that $F$ is an isomorphism of vector spaces.  To see that it is also a homeomorphism, first check that $F$ is continuous.  Let $\lambda \mapsto y^{\lambda}\in \ell ^{\infty}$ converge to $y\in \ell ^{\infty}$ and let $B\subseteq \ell ^p$ be $\ell ^1$-bounded.  Then, there is some constant $C$ such that $\| x\| _1\leq C$ for all $x\in B$.  Hence,
$$
\sum _{m\in \mathbb{N}}|(y_m^{\lambda}-y_m)x_m|\leq \| y^{\lambda}-y\| _{\infty}\| x\| _1\leq C\| y^{\lambda}-y\| _{\infty},
$$
and so $\lambda \mapsto y^{\lambda}$ converges to $y$ uniformly on $B$.
Conversely, suppose that $\lambda \mapsto y^{\lambda}$ converges to $y$ uniformly on every $\ell ^1$-bounded subset of $\ell ^p$.  Take $B:=\{ \mathbf{e}_m:m\in \mathbb{N}\}$, where $\mathbf{e}_m\in \ell ^p$ is of course the sequence which is identically $0$ except for a $1$ at the $m^{\text{th}}$ index.  This is certainly $\ell ^1$-bounded, and hence for every $\varepsilon >0$ there is some $\lambda _0$ such that whenever $\lambda \geq \lambda _0$ it follows that
$$
\left| (y^{\lambda}-y)(\mathbf{e}_m)\right| =\left| y^{\lambda}_m-y_m\right| <\varepsilon .
$$
for all $m\in \mathbb{N}$.  That is, $\lambda \mapsto y^{\lambda}\in \ell ^{\infty}$ converges to $y\in \ell ^{\infty}$ with respect to the $\ell ^{\infty}$ norm.
