inequality near $0$ with arbitrary norm

Consider $a(x,y)= 2 x y - x^3+ x y^3$ and $b(x,y) = || (x,y)||^2$ for arbitrary norm. I want show that: $$|a(x,y)| \leq C |b(x,y)|$$ for $||(x,y)|| < \delta$ for $C, \delta >0$

Is there a general strategy since I do not know which norm I have to deal with?

Solution 1:

When speaking about equivalent norms, in a sense that if $n_1(x)$ and $n_2(x)$ are different norms of $x$, we must have positive constants $c_1,c_2$ such that $$c_1n_1(x)\le n_2(x)\le c_2n_1(x),$$then we can reduce the problem to the Euclidean norm and proceed.

A good way to tackle these types of problems, is to work with bounds and inequalities rather than being too much strict. More specifically, start with showing that $$ |a(x,y)|\le |2xy|+|x^3|+|xy^3|\le Cb(x,y) $$ for some $C>0$ given $\delta>0$. The confidence in such an argument is that $x$ and $y$ cannot jump high and they always remain below a threshold ($\delta$). We continue as $$ { |2xy|<x^2+y^2 \\ |x^3|<\delta x^2 \\ |xy^3|<\delta^2 y^2 }. $$Hence by defining $C=1+\max\{\delta,\delta^2\}$, we complete the proof for $2$-norm and any other norm equivalent to it.