# Is the triangle inequality too strong?

A norm $\Vert\cdot\Vert$ on a vector space $V$ is required to satisfy the triangle inequality: $\Vert x+y\Vert\leqslant\Vert x\Vert+\Vert y\Vert.$ The function of this inequality is to ensure that mapping $f:V^2\to V,\ (x,y)\mapsto x+y$ is continuous. But why don't we simply require that $\displaystyle\lim_{(x,y)\to (0,0)} \Vert x+y\Vert=0,$ instead of the subadditivity? Or, can we repalce the triangle inequality with $\Vert x+y\Vert\leqslant 2\Vert x\Vert +3\Vert y\Vert,$ or something like that?

Question: What's the irreplaceability of triangle inequality in the normed vector space theory?

• en.wikipedia.org/wiki/Quasinorm May 5, 2018 at 5:05
• @vadim123:Thanks a lot! Shamed of my ignorance... :( May 5, 2018 at 5:10

The function of this inequality is not to have "addition" continuous. The triangular inequality is also required for metric spaces, such as $\mathbb{Q}$ with the discrete metric. I do not think that people are actually interested in discussing continuity of the "+" function.
The triangular inequality is based on a much more basic idea: The norm should represent the length of a vector $x+y$. When you are not measuring "straight" from the beginning $0$ to the end $x+y$, but taking some detour via $x$, you should at least not get a shorter measurement.
• This situation can occur, for instance in an $L^p$ space, with 0 < p < 1. There is also the concept of Quasinorm which is a generalisation of a norm that does not fulfil subadditivity - but a similar property. May 5, 2018 at 5:43
As I understand it the triangle inequality derives from metric spaces: "Visually" spoken, the direct connection between two points $A$ and $B$ should never be longer than the sum of the two distances $\overline{AC}$ and $\overline{CB}$ for any other point $C$.