I am given a function $\|\cdot\|: \Bbb R^2 \rightarrow \Bbb R$ that satisfies all the conditions of a norm except the triangle inequality. And let:
$$ B = \{\mathbf v \in \Bbb R^2 \mid \|\mathbf v\| \leq 1 \}$$
I am asked to show that if $B$ is convex, namely if $\mathbf{v},\mathbf{w} \in B, \: t\mathbf v + \left(1-t\right)\mathbf w \in B \:\:\: \forall t \in \left[0,1\right]$, then $\|\cdot\| $ satisfies the triangle inequality.
I have thought that maybe I can prove this by contradiction, assuming that there exists some pair of vectors that do not satisfy the triangle inequality under this function. However, I need to prove the result that:
$$ \|\mathbf v + \mathbf w\| > \|\mathbf v\| + \|\mathbf w\| \Rightarrow \|l\mathbf v + k\mathbf w\| > |l| \|\mathbf v\| + |k|\|\mathbf w \| \:\: \forall l,k \in \Bbb R$$
And I cannot seem to do this. If this result is true than I can consider the unit versions of the vectors and show that convexity forces the triangle inequality and thus, a contradiction. However, I cannot prove this result.
Would my approach work? Is there perhaps a shorter, more concise and neater approach that would achieve the desired proof? Any advice and help you may be able to offer would be greatly appreciated!
Best Answer
I assume you mean that $\|\cdot\|$ satisfies all the conditions for a norm, except for the triangle inequality.
Since $\|\lambda x\|=|\lambda|\|x\|$ for all scalars $\lambda$, it follows that $$\frac{x}{\|x\|}\in B$$ for non-zero $x$. But then: $$\frac{v+w}{\|v\|+\|w\|}=\frac{\|v\|}{\|v\|+\|w\|}\frac{v}{\|v\|}+\frac{\|w\|}{\|v\|+\|w\|}\frac{w}{\|w\|}\in B$$ since $B$ is convex. Therefore, by the definition of $B$: $$\frac{\|v+w\|}{\|v\|+\|w\|}\le 1$$