Let $E$ be a normed space and $u,v \in E$ so that $\|u+v\|=\|u\|+\|v\|$. Prove that $\forall s,t \geqslant 0$ we have that $\|su+tv\|=s\|u\|+t\|v\|$
Note: This is homework, so I'm looking for hint(s) only.
(some of) My attempts:
The trivial cases:
- $u =0$ or $v = 0$.
- $\exists \lambda > 0$ so that $u=\lambda v$.
So we can assume that neither of those is true here.
It's easy to see that it's equivalent to proving that $\forall \lambda \geqslant 0$ we have that $\|u+\lambda v\|=\|u\|+\lambda \|v\|$, because if $s > 0$: $$s\|u+\lambda v\|=\|su+s\lambda\|=s\|u\|+s\lambda \|vu\|$$ So we can pick $\lambda=\frac{t}{s}$.
I tried to take the derivative of $\lambda \mapsto \|u+\lambda v\|$, but I could only prove that $$\left|\frac{\|u+(\lambda + h)v\|-\|u+\lambda v\|}{h}\right|\leqslant \frac{\|hv\|}{|h|}=\|v\|$$ But it's not enough to conclude that it's equals to $\|v\||$ in the limit $h \to 0$.
Another idea was to assume that $\exists \alpha > 0$ so that $\|u+\alpha v\| < \|u\| + \alpha \|vu\|$ and prove that $\|u+v\| \neq \|u\|+\|vu\|$, but it was not successful.
- I was thinking about proving it for $\lambda \in \mathbb{Q}_+$ and use some kind of density argument, but it seems to be as hard as the $\mathbb{R}_+$ case.