Showing that $d:P_{3} \times P_{3} \rightarrow \mathbb{R}$ is not a metric. The metric is defined like so. $d:P_{3} \times P_{3} \rightarrow \mathbb{R}$, where $P_{3}$ is the set of all polynomials of degree less than or equal to $3$.
$d(p,q) = |(0)− (0)|+|(1)− (1)|+|(2)−(2)|$. I'm trying to show that this isn't a metric. My best attempt is that it's not a metric due to the fact that it's not definite. That is, that $d(p,q)=0 \nRightarrow p = q.$ But I'm not sure exactly how to prove this claim. The best I could come up with is that if $(ax^3+bx^2+cx+d) - (a^{'}x^3+b^{'}x^2+c^{'}x+d^{'})=0,$ it could just be the case that $d=d^{'}$, while the rest of the coefficients are not equal. So $p\neq q.$
But then I also have to prove that $d:P_{3} \times P_{3} \rightarrow \mathbb{K}$, $\delta(p, q) = \sum_{j=0}^N|()−()|$ is a metric for $N > 3$, where $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$. I'm not really sure where to start. Any help is appreciated.
 A: A cubic polynomial has arbitrary values on any four points, and therefore two such polynomials can agree on a triple of points, so you can have two different polynomials which agree on $0, 1, 2.$ The same reasoning shows that your second metric is not indefinite. It quite obviously satisfies the other axioms of a metric (the only one that is not completely trivial is the triangle inequality, but that is not hard either).
A: If we want the two polynomials to have a non-zero cubic term we can use $$p(x)=3x^3-5x^2+5x$$ and $$q(x)=x^3+x^2+x$$
$p(0)=q(0)=0\\p(1)=q(1)=3\\p(2)=q(2)=14$
In fact, to get this and others systematically, we notice that if a polynomial $p(x)=ax^3+bx^2+cx+d$, then to equate the values of $0$, $1$ and $2$ to another polynomial we need to have $(a,b,c)$ for each polynomial lie on the two planes $$x+y+z=k_1$$ and $$8x+4y+2z=k_2$$ for some constants $k_1,k_2$.
Taking the cross product $$(8\hat i+4\hat j+2\hat k)=(\hat i+\hat j+\hat k)=2\hat i-6\hat j+4\hat k$$ gives us that any two points on a line parallel to the cross product will give us a set of coefficients that work. Thus $(1,1,1)+(2,-6,4)=(3,-5,5)$ works as above but the polynomial $$r(x)=5x^3-11x^2+9x$$ also works.
