Norm defined on a Real Vector Space of Polynomials

Question

Consider the real vector space $V$ of polynomials of degree less than or equal to $d$. For $p \in V$ a norm is defined as $\|p\|_k = \max \{|p(0)|,|p^{(1)}(0)|, \cdots, |p^{(k)}(0)|\} $, where $p^{(i)}(0)$ is the $i^{th}$ derivative of $p$ evaluated at 0. Then $\|p\|_k $ defines a norm if and only if
A. $k \geq d-1$
B. $k \lt d$
C. $k \geq d$
D. $k \lt d-1$

So, far option (B) and (D) are clear to me. For option B, I have taken $d=2$ , and a polynomial $p(x)=x^2$, so it is clear to me that $\|p\|_k =0$ but $p(x)$ is non-zero, which fails the first property of being a norm.So, it fails for $k \lt d$ that is for $k \leq d-1$. Hence, for option (D).I am not getting any idea for (A)and (C).

Answer 1

Suppose that $P(x)=a_0+..+a_dx^d$, $\|P\|_k=max(|a_0|,|a_1|,...,k!|a_k|)$, so it is a norm if $k\geq d$. If $k<d$, $\|x^d\|=0$, it is not a norm which generalize your argument. If $k>d$, $\|P\|_k=max(|a_0|,|a_1|,...,d!|a_d|)$. Norm $\|P\|_k=0$ if and only if $P=0$. You can prove the others properties.