How do I prove that the following map is onto? We have the map $\phi :\ell q \to \ell p'$ defined by $(y_n)=y \mapsto f_y$. where $$f_y(x)=\sum_{n=1}^ \infty x_n y_n, \quad \quad x=(x_n) \in \ell_p.$$ I know $(e_k)$ is a Schauder basis for $\ell_p$, so each $x \in \ell_p$ has a unique representation $\sum_{n=1}^{\infty} x_n e_n$. To prove the map surjective, I'm picking $g\in \ell_p'$ and then $$g(x)=\sum_{n=1}^{\infty}x_ny_n, \quad \quad g(e_k)= y_k.$$
Now I don't see how to show $(y_n) \in \ell_q$, which I think will prove that $g=g_y$?
 A: Let $\varphi\in \ell'_p$ and $\{e_n\}$ the standard Schauder basis of $\ell_p$, and $y_n=\varphi(e_n)$. We shall show that $y=(y_1,\ldots,y_n,\ldots)\in \ell_q$ and $\varphi(x)=\langle x,y\rangle$, for all $x\in\ell_p$.
Let first $\varphi_n(x)=\sum_{k=1}^n x_ky_k$, where $x=(x_1,\ldots,x_n,\ldots)\in\ell_p$. Clearly,
$$
\Big(\lvert y_1\rvert^q+\cdots+\lvert y_n\rvert^q\Big)^{1/q}=\|\varphi_n\|\le \|\varphi\|,
$$
and hence $y\in\ell^q$, and $\|y\|_q\le \|\varphi\|$.
Next, define $P_n$ to be the projection from $\ell_p$ to the span of $\{e_1,\ldots,e_n\}$. We have that $\varphi_n(x)=\varphi(P_nx)$ and hence
for every $x\in\ell_p$
$$
\lvert\varphi_n(x)-\varphi(x)\rvert=\lvert\varphi(P_nx)-\varphi(x)\rvert
=\lvert\varphi(P_nx-x)\rvert\le \|\varphi\|\|P_nx-x\|_p.
$$
Next observe that $\|P_nx-x\|_p\to 0$ as $n\to\infty$, for all $x\in\ell_p$, and also 
$$
\lvert\varphi_n(x)-\langle x,y\rangle\rvert\le\sum_{k>n}\lvert x_k\rvert\lvert y_k\rvert\to 0.
$$
Thus $\varphi(x)=\langle x,y\rangle.$
A: Assuming $p > 1$. For any sequence with finite non-zero terms $(a_1, a_2,\dots, a_n, 0, 0, \dots)$ in $\ell_p$, $$a = \sum_{i=1}^n a_i e_i.$$ $$g(a)=\sum_{i=1}^n a_iy_i.$$
Let $x_i = |y_i|^q y_i^{-1}$ for $i = 1, \dots, n$ and $= 0$ for $i > n$ or $y_i = 0$.
$$\|x\|_p = (\sum_{i=1}^n |y_i|^q)^{1/p},$$ and $$g(x) = \sum_{i=1}^n |y_i|^q.$$
$$\infty > \|g\|_{\ell_p'} = \sup_{0\not= a \in \ell_p} \frac{|g(a)|}{\|a\|_p}\ge \frac{|g(x)|}{\|x\|_p} = \frac{\sum_{i=1}^n |y_i|^q}{\sum_{i=1}^n (|y_i|^q)^{1/p}} \to \sum_{i=1}^{\infty} (|y_i|^q)^{1/q} \, \text{as} \, n \to \infty. $$
so that $(y_n) \in \ell_q$ and $\phi (y) = g$.
Source: http://www.maths.tcd.ie/~richardt/321/321-ch3.pdf
(Full disclosure: Something that is bothering me still, it is not clear if $g = g_y$ since the sum on $g$ has range $i=1, \dots, n$ and not $i=1, \dots, \infty$, and it would help if someone here can comment on this.)
