Definition of a support point:
Let $X$ be a vector space over $\mathbb{R}, P\subseteq X$ is a set, a point $\xi\in P$ is said to be a support point of $P$ if there is some hyperplane $H=[\psi=\alpha]$ such that $\xi\in H,P\not\subseteq H$ and $P$ lies in one of the half space determined by $H,$ namely $$\psi(\xi)=\alpha,\psi\big{|}_P\geq \alpha~\text{and}~\exists x_0\in P,\psi(x_0)>\alpha$$ here $0\ne\psi\in X^{\prime}$ (the algebraic dual space of $X$). And such a hyperplane is said to support $P$ at $\xi.$
Now let $X=\ell^{p}$ with $1\leq p<+\infty, P\subseteq X$ is the positive wedge defined by $$P=\left\{x=(x_n)\in X:x_n\geq 0\right\}$$ let $\xi=(\xi_n)\in P$ be a point with positive coordinates, namely $\xi_n> 0.$ I want to show that $\xi$ is not a support point of $P.$
Here is my work:
Suppose by contradiction that there is some hyperplane $H=[\psi=\alpha]$ that supports $P$ at $\xi.$ If $\alpha>0,$ note that $\frac{\xi}{2}\in P,$ so we have $\alpha\leq \psi(\frac{\xi}{2})=\frac{1}{2}\psi(\xi)=\frac{1}{2}\alpha,$ which is absurd! Thus $\alpha\leq 0,$ similarly we can get $\alpha\geq 0$ and consequently $\alpha=0.$
If $\psi$ is a bounded linear functional, by Riesz's representation there is some $\eta=(\eta_n)\in \ell^{q}$ ($1/p+1/q=1$) such that $\psi(x)=\sum_{n=1}^{\infty}x_n\eta_n$ for all $x=(x_n)\in X.$ Since $e_n=(0,\dots,0,1,0,\dots)\in P$ ($1$ is in the n-th component) and $\psi\big{|}_P\geq 0,$ we have $\eta_n=\psi(e_n)\geq 0,\forall n.$ So $0=\psi(\xi)=\sum_{n=1}^{\infty}\xi_n\eta_n\geq 0,$ the equality sign holds iff $\eta_n=0,\forall n$ since $\xi_n>0,$ we derive that $\eta_n=0,\Rightarrow \psi=0,$ which is a contradiction!
Hence it remains to consider the case when $\alpha=0$ and $\psi$ is a unbounded linear functional.
I also make some efforts in this case but it seems to be of less help: Because $P\not\subseteq H,$ we can find some $y=(y_n)\in P$ such that $\psi(y)>0,$ if we can find some $\lambda>0$ so that $\xi-\lambda y\in P,$ then by linearity we have $0\leq \psi(\xi-\lambda y)=-\lambda\psi(y)<0,$ which leads to a contradiction and completes the proof. But the problem is that such a $\lambda$ may not exists (for example if $\xi_n/y_n\to 0,y_n>0$ and such a $\lambda$ exists, then $\xi-\lambda y\in P$ implies $\xi_n-\lambda y_n\geq 0$ or equivalently $\xi_n/y_n\geq \lambda,$ let $n\to\infty$ we can see $\lambda=0$.).
Is there anyone who can help me?
The functional $\psi$ is positive (since $\alpha = 0$) and positive functionals on $\ell^p$ (and more generally on Banach lattices) are automatically continuous. See for instance Theorem II.5.3 on page 84 of Helmut H. Schaefer's book "Banach Lattices and Positive Operators", 1974 (zbMATH link).
(One might remark, by the way, the Holmes' definition of support points is really very algebraic, probably due to the fact that the book seems to be set mainly a vector space setting. When working on Banach spaces one would typically prefer to define support points by means of bounded linear functionals.)