When is this norm defined? This question was part of my linear algebra assignment and I was unable to solv eit. SO, I am asking you for help.

Consider the vector space V of real polynomials of degree less than or equal to n. Fix distinct real numbers $a_0 , …, a_k$ . For $p \in V$ define $max${${|p(a_j)| : 0\leq j \leq k}$} . It defines a  norm on V if:

A only if k<n
B only if $k\geq n$
C if $k+1 \leq n $
D if $k\geq n+1$
I tried thinking on which result / reasoning I should use but I am clueless on when the norm will not be defined .
Can you please help me through it?
 A: Answer. $B$: Only if $k\ge n$.
First, we study the case $k=n$.
If $p(x)=c_0+c_1x+\cdots+c_k x^k$, and
$$
p(a_j)=b_j, \quad j=0,1,\ldots,k, 
$$
then the above is a $(k+1)\times (k+1)$ linear system
$$
c_0+c_1 a_j+\cdots+c_n a_j^k=b_j, \quad j=0,1,\ldots,k, 
$$
with unknowns the $c_j$'s. The determinant of the matrix $M$ the system is the Vandermonde one, hence non-vanishing, and thus the system possesses a unique solution
$$
c=(c_j)=M^{-1}b=M^{-1}(b_j).
$$
Hence, if the $b_j=p(a_j)$ all vanish, so does $p$.
Clearly, $\|p\|_V=\max\{|c_j|\}$ defines a norm in $V$.
Also,
$$
\|p\|_V=\|c\|\le \|M^{-1}\|\cdot\|b\|
$$
where
$$
\|c\|=\max_{j=1,\ldots,n}|c_j|, \quad \|A\|=\sup_{\|x\|=1}\|Ax\|.
$$
and since $Mb=c$
$$
\|b\|\le \|M\|\cdot\|c\|=\|p\|_V
$$
Thus $\|b\|=\max_{j=0,1,\ldots,k}\{|p(a_j)|\}$ is a norm equivalent to $\|p\|_V$. In particular, if
$\max_{j=0,1,\ldots,k}\{|p(a_j)|\}=0$, then $p\equiv =0$.
If $k>n$, then it is still a norm, since
$$
\max_{j=1,\ldots,n}\{|p(a_j)|\}=0\quad\Longrightarrow\quad
\max_{j=1,\ldots,k}\{|p(a_j)|\}=0\quad\Longrightarrow\quad p\equiv 0.
$$
If $k<n$, one can then $p(x)=(x-a_0)(x-a_1)\cdots(x-a_k)$ vanishes at the $a_j$'s and hence $\max_{j=1,\ldots,k}\{|p(a_j)|\}=0$, while it is a non-vanishing polynomial of degree $k+1\le n$.
A: Let $n=2$. If $k=1$, let $\|.\|$ be the corresponding defined norm.
If $a_0=1, a_1=-1$, then we have $\|x^2-1\|=0$, even thought $x^2-1 \ne 0$.
In fact, if $n=2$ and $k=1$, then we have $\|(x-a_0)(x-a_1)\|=0$.
In general, if $k+1 \le n$, then we have $\|\prod_{i=0}^k (x-a_i) \|=0$ and $\prod_{i=0}^k (x-a_i)$ is of degree $k+1$ and non-zero.
Hence we need $k \ge n$ due to the positive definite requirement.
Now to verify sub-additivity,
\begin{align}
\|f + g\| &= \max_{0 \le i \le k} \{|(f+g)(a_i)| \} \le \max_{0 \le i \le k} \{|f(a_i)| + |g(a_i)| \}  \le \|f\| +  \|g\|
\end{align}
Verification of absolutely homogeneous is direct.
As an exercise, you might like to answer the question when is it a seminorm.
A: I think the answer should be both B and D.
Since is $D \subset B.$
