I have a question from Brezis' Functional Analysis- Sobolev Spaces and Partial Differential Equations.
In the proof of Thm 4.11 (Riesz representation theorem) for $L^p$ spaces ($1 < p < \infty$), it says the following.
We consider the operator $T : L^{p'} \to (L^p)^*$ defined by $\langle Tu, f\rangle = \int uf$, $\forall u \in L^{p'}$ , $\forall f \in L^p$. The argument used in the proof of Theorem 4.10 shows that $$\Vert Tu \Vert _{(L^p)^*} = \Vert u\Vert_{p'} \quad \forall u \in L^{p'}.$$ We claim that $T$ is surjective. Indeed, let $E = T(L^{p'})$. Since $E$ is a closed subspace, it suffices to prove that $E$ is dense in $(L^p)^*$. Let $h \in (L^p)^{**}$ satisfy $\langle h, Tu \rangle = 0$ $ \forall u \in L^{p'}$. Since $L^p$ is reflexive, $h \in L^p$, and satisfies $\int uh = 0$ $\forall u \in L^{p'}$. Choosing $u = \vert h\vert^{p-2}h$, we see that $h = 0 $.
Here, I think he says if any continuous linear functional on $ (L^p)^*$ which vanishes on $\overline{Y} \subset (L^p)^*$ vanishes on the whole $ (L^p)^*$, then $\overline{Y} = (L^p)^*$. To prove this, I tried proof by contradiction. In other words, if $\overline{Y} \neq (L^p)^*$, then there is a linear continuous functional which vanishes on $\overline{Y}$, but is not identically zero. To do that, I chose $x_0 \in (L^p)^* \setminus \overline{Y}$ and constructed the function $\gamma: \overline{Y} \oplus \mathbb{R}x\to \mathbb{R}$ by defining $\gamma (y+tx) = t$. If this functional is continuous, then I can extend this to $(L^p)^*$ by Hahn-Banach. However, I cannot show $\gamma$ is continuous. Could you help me with this? I'd appreciate any other method of tackling this. Thank you very much.