$V=P_n(\mathbb{R})$ be the vector space of all polynomials with degree $\le n$. I need to know Which of the followings are norms on $V.$
$\forall p\in V$
1.$\|p\|^2=|p(1)|^2+\dots+|p(n+1)|^2$
2.$\|p\|=\sup_{t\in [0,1]}|p(t)|$
3.$\|p\|=\int_{0}^{1}|p(t)|dt$
4.$\|p\|=\sup_{t\in [0,1]}|p'(t)|$
for the $1$, $\|p\|=0\Rightarrow p\equiv 0$, also $\|\lambda p\|=|\lambda|\|p\|$,but I think $\|p+q\|\le \|p\|+\|q\|$ doesn't hold here, but I am not sure about the proof.
$2$ is our wellknown sup norm so it indeed a norm on $V$
$3$ is also a norm I am sure.
I can not say about $4$. Thanks for correcting me
Let $p=c$, $c$ is a non-zero constant. Degree of $p$ is $0$ and $p\neq 0$ but $\|p\|=\sup_{x\in [0,1]}|p'(x)|=0$ so 4 is not a norm.
EDIT: Thanks for warnings. Degree of $p$ is $n+1$ so it is not in $P_n(\mathbb R)$. If $p=0$ then trivially $\|p\|=0$. If $\|p\|=0$ then $|p(1)|=|p(2)|=\cdots=|p(n+1)|=0$ so the polynomial $p$ has roots at the points $1,2,\ldots,n+1$. This is impossible because $\deg(p)\le n$ so $p$ must be $0$. Absolute homogeneity of the function in (1) can be seen easily and the triangle inequality of (1) can be shown using Cauchy-Schwarz inequality. Thus the function in (1) is a norm.