We define the absolute value in $\mathbb{Q}$ as an application $||\, \cdot \, || : \mathbb{Q}\rightarrow [0,\infty )$ that fulfills the properties:
- $||x||=0$ if and only if $x=0$.
- $||xy||=||x||\, ||y||$.
- $||x+y||\leq ||x||+||y||.$
We have to show that if for $n\in \mathbb{Z}$ we have $||n||\leq 1$, then it is fulfilled that for all $x,y\in \mathbb{Q}$ we have $$||x+y||\leq \max (||x||, ||y||).$$
My attempt is as follows:
$$||x+y||^N =\prod_{i=1}^N ||x+y|| =||(x+y)^N||=|| \sum_{k=0}^N {N\choose k}x^{N-k}y^k|| \leq \sum_{k=0}^N ||{N\choose k}x^{N-k}y^k||$$
Can you help me doing this proof?
Thanks
Let $(x,y)\in\mathbb{Q}^2$, you have $$ \|x+y\|^n=\|(x+y)^n|\|=\left\|\sum_{k=0}^n{\binom{n}{k}x^k y^{n-k}} \right\|\leqslant\sum_{k=0}^n{\underbrace{\left\|\binom{n}{k}\right\|}_{\leqslant 1}\|x^k y^{n-k}\|}\leqslant\sum_{k=0}^n{\|x\|^k\|y\|^{n-k}} $$ Since $\|x\|\leqslant\|\max(x,y)\|$ and $\|y\|\leqslant \|\max(x,y)\|$ you then have $$\|x+y\|^n\leqslant\sum_{k=0}^n{\|\max(x,y)\|^k\|\max(x,y)\|^{n-k}}=(n+1)\|\max(x,y)\|^n $$ Hence $$ \|x+y\|\leqslant (n+1)^{\frac{1}{n}}\|\max(x,y)\|\underset{n\rightarrow +\infty}{\longrightarrow}\|\max(x,y)\| $$