I'm reading about the modulus of continuity. Consider on the space of polynomials of degree $\leq n$ a number:
$$ \omega\big(p, [-1,1], \delta\big) = \sup_{|x_1 - x_2| \leq \delta} \big| p(x_1) - p(x_2) \big|$$
Given the interval $I = [-1,1]$ and a box $\delta > 0$ and is this thing a norm on the space of polynomials
$$ P_n = \{ p(x) : \deg p \leq n \}$$
We need to include lower degree polynomials in order to count for examples like $x^3 - (x^3 - 1) = 1$ and $\deg 1 = 0$. The axioms of a norm are as follows:
- $||\mathbf{0}|| = 0$ and every other vector has positive length $||\vec{v}|| > 0$ for all $\vec{v} \in V$.
- Multiplying a vector changes it's length but not it's direction $||a\, \vec{v}|| = |a| \, ||\vec{v}||$.
- The triangle inequality holds: $ ||\vec{v} + \vec{w}|| \leq ||\vec{v}|| + ||\vec{w}||$.
So this is already not a norm since I can find many functions with zero modulus of continuity... the constant funtions:
$$ \omega \big(1, [-1,1], \delta \big) = \sup |1 - 1| = 0 $$
I'm guessing not 100% sure the only polynomials with nonzero polynomials with $\omega(\delta) = 0$ are the constant functions $p(x) = c \cdot 1$ However this looks mostly like the sup norm and the be two proposals:
$\omega(p,\delta)$ is a norm on the quotient vector space $P_n / \langle 1 \rangle = [polynomials] / [constants]$.
If I add the sup norm we can get another norm: $||p|| + \omega(p, \delta)$.
I beleive $\omega$ satisfies scalar multiplication and triangle equality. Perhaps both or neither are norms on their respective vector spaces.
It is not difficult to show that $n(p)= \omega(p,[-1,1]\delta)$ satisfies $n(\lambda p) = |\lambda| n(p)$ and $n(p+q) \le n(p)+n(q)$.
Hence $\|p\| = n(p)+\|p\|_\infty$ satisfies the homogeneity and subadditivity requirements of a norm.
All that remains is to show that $\|p\|= 0$ implies $p=0$, but this follows from $\|p\|_\infty = 0$.