Riesz representation Theorem by Brezis I don't understand the following proof:
Theorem: Let $1<p<\infty$ and let $\phi\in(L^p)^*$. Then there exists a unique function $u\in L^q$ such that
$\langle\phi\,,f\rangle=\int uf$ for all $f\in L^p.$
Moreover, $||u||_{q}=||\phi||_{(L^p)^*}$ and $\phi$ is surjective.
Proof: Consider $T:L^q\rightarrow(L^p)^*$ defined by 
$\langle Tu\,,f\rangle=\int uf$. For all $u\in L^q$ and all $f\in L^p$.
(I understand most of the proof except the surjectivity, it says the following)
Let $E=T(L^q)$ since E is a closed subspace it suffices to prove that $E$ is dense in $(L^p)^*$. Let $h\in(L^p)^{**}=L^p$ (because we have already proved that for $1<p<\infty$, $L^p$ is reflexive) such that 
$\langle Tu\,,h\rangle=\int uh=0$ for all $u\in L^p$. Choosing $u=|h|^{p-2}h$ we see that $h=0$. 
My question is how does this prove surjectivity or density? Thanks in advance.
 A: For completeness recall that any $f \in L^p$ induces a linear function $\Phi_f \in (L^p)^{**}$ via the action
$$\langle \Phi_f, u \rangle = \langle u, f \rangle, \quad u \in (L^p)^*.$$
Since $L^p$ is reflexive (which Brezis proves independently of the Riesz Representation Theorem), every $\Phi \in (L^p)^{**}$ equals some $\Phi_f$.
Suppose $E$ is not dense in $(L^p)^*$. According to the Hahn-Banach theorem, there exists a nonzero linear function $\Phi_f \in (L^p)^{**}$ that vanishes on $E$. In particular $$0 = \langle \Phi_f, Tu \rangle = \langle Tu,f \rangle = \int uf$$for all $u \in L^q$. Since $f \in L^p$, you have that $h(x) = f(x)|f(x)|^{p-2} \in L^q$, because
$$\int |h(x)|^q = \int |f(x)|^{q(p-1)} = \int |f(x)|^p.$$
Consequently
$$0 = \langle Th,f \rangle = \int f(x) |f(x)|^{p-2} f(x) = \int |f(x)|^p.$$
This forces $f = 0$ almost everywhere, so that $f$ is the zero element of $L^p$ and  thus$$\langle \Phi_f,u \rangle = \langle u,f \rangle = 0$$
for all $u \in L^q$.
This contradicts the supposition that $\Phi_f$ was a nonzero functional. Thus $E$ is dense. Since $E \subset (L^p)^*$ is closed, this forces $E = (L^p)^*$.
