For concatenation functions such as $f(\vec{p}) =(\vec{p},\vec{p})$ or $g(\vec{p}) =(\vec{p},\vec{1}-\vec{p})$. where $\vec{p}, \vec{1}-\vec{p} \in P \subseteq [0,1]^n $. What's the property of it? is it convex? how should I approach it?
What my context is that $\vec{p} \in P$ where $P$ is a convex set. And I have a convex function $l$ where ${\displaystyle l\colon \,P\times P\mapsto \mathbb {R}}$.
For the composed function $l(g(\vec{p}))$ can i claim that
$ {\displaystyle {\underset {\vec{x}\in P}{\operatorname {arg\,max} }}\,l(g(\vec{x}))\in \text{closure of} convexhull(P) } $ ?
Vector-valued functions do not have a notion of convexity, so no.
By definition, if you take the argmax over $\vec{x} \in P$, you will get out an element $\vec{x}_0 \in P$. Since $P$ is convex, it is equal to its convex hull and the result trivially holds.
Perhaps you want to know whether $l(f(\vec{p}))$ or $l(g(\vec{p}))$ are convex. That is, does it hold that $$l(f(t\vec{p}_1+(1-t)\vec{p}_2)) \leq t l(f(\vec{p}_1)) + (1-t) l(f(\vec{p}_2)) \forall t \in [0,1]$$ and similarly for $g$. For $f$, notice that $$f(t\vec{p}_1+(1-t)\vec{p}_2) = (t\vec{p}_1+(1-t)\vec{p}_2,t\vec{p}_1+(1-t)\vec{p}_2)= t(\vec{p}_1,\vec{p}_1)+(1-t)(\vec{p}_2,\vec{p}_2).$$ Now use the convexity of $l$.
For $g$ it is a little trickier, but you can show that \begin{align*} g(t\vec{p}_1,(1-t)\vec{p}_2)&=(t\vec{p}_1+(1-t)\vec{p}_2, \vec{1}-(t\vec{p}_1+(1-t)\vec{p}_2))\\ &= (t\vec{p}_1+(1-t)\vec{p}_2, t (\vec{1}-\vec{p}_1) + (1-t)(\vec{1}-\vec{p}_2))\\ &= t(\vec{p}_1, \vec{1}-\vec{p}_1) + (1-t) (\vec{p}_2, \vec{1}-\vec{p}_2). \end{align*} Now again use the convexity of $l$.