General Triangle Inequality Conditions for Equality

General Triangle Inequality Conditions for Equality
$\begingroup$

I am trying to find necessary and sufficient conditions for the triangle equality to be satisfied:

$$|x_1 + x_2 + \cdots + x_n| = |x_1| + |x_2| + \cdots + |x_n|$$

where $x_i$ are vectors in $\mathbb R^n$.

I am guessing the condition is analogous to the $|x+y|\le|x|+|y|$ case, namely that $x_1, x_2, \ldots x_n$ are a linearly dependent set. I can prove this is necessary, but not that it is sufficient, which it may well not be.

$\endgroup$
4

1 Answer

$\begingroup$

Suppose that none of $x_1,x_2,...,x_{n+1}$ is zero, observe that when $n=1$ $$|x_1+x_2|=|x_1|+|x_2| $$ if and only if $\exists c_1>0$ such that $x_2=c_1x_1$.

By triangle inequality $|x_1+...+x_n+x_{n+1}|\le |x_1+...+x_n|+|x_{n+1}|$ so if $$|x_1+...+x_n+x_{n+1}|= |x_1|+...+|x_n|+|x_{n+1}|$$ ,we'd have $|x_1|+...+|x_n|\le|x_1+...+x_n|$ which implies that $$|x_1|+...+|x_n|=|x_1+...+x_n|$$

Proceed inductively, assume that $$|x_1+...+x_n|=|x_1|+...+|x_n| $$ if and only if $\exists c_1,...,c_{n-1}>0$ such that $x_{i+1}=c_ix_1$ whenever $i=1,2,...,n-1$. If we have $$|x_1+...+x_n+x_{n+1}|=|x_1|+...+|x_n|+|x_{n+1}| $$ then $$|x_1+...+x_n+x_{n+1}|=|x_1+...+x_n|+|x_{n+1}| $$ so $\exists c^*_n>0$ such that $x_{n+1}=c^*_n(x_1+...+x_n)=c^*_n(1+c_1+...+c_{n-1})x_1$. By letting $c_n=c^*_n(1+c_1+...+c_{n-1})$, we arrive the conclusion that

$|x_1+...+x_n+x_{n+1}|=|x_1|+...+|x_n|+|x_{n+1}|$

if and only if $\exists c_1,...,c_n>0$ such that $x_{i+1}=c_ix_1$.

The case where some of $x_i=0$ can be tackled analogously with some caution.

$\endgroup$
5

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

James H. Sterling
Author

James H. Sterling

James Sterling reports on renewable energy developments, climate policy, ecological conservation, and green tech innovations around the globe.