# strict subadditivity of the norm in uniformly convex Banach Spaces.

So here is my question,

I would like to prove the following,

Let $(X,||\cdot||)$ be a uniformly convex Banach Space. Then the norm is strictly subadditive i.e $\forall x,y\in X\backslash\{0\}:||x+y||<||x||+||y||$

I started my prove by considering two elements satisfying $||x+y||=||x||+||y||$. By using all properties of a unifomly convex space it should follow that $x=y=0$ but I always failed.

Can someone give me a hint gow to proceed?

Thanks!

• You may want $x$ to not be a scalar multiple of $y$. – David Mitra May 7 '14 at 13:24
• @DavidMitra So my definition for strict subadditivity is wrong? Should it be, if for $x,y\in X$ there exists $\lambda\in \mathbb K$ such that $x=\lambda y$ then $x=y=0$? – Thorben May 7 '14 at 13:53

If $x$ is not a multiple of $y$, $\Vert y\Vert\ge\Vert x\Vert>0$, and $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$, then $$\biggl\Vert {x\over\Vert x\Vert}+ {y\over\Vert y\Vert} \biggr\Vert \ge \biggl\Vert {x\over\Vert x\Vert}+ {y\over\Vert x\Vert} \biggr\Vert - \biggl\Vert {y\over\Vert x\Vert}- {y\over\Vert y\Vert} \biggr\Vert ={{\Vert x\Vert +\Vert y\Vert}\over \Vert x\Vert} -\Vert y\Vert\Bigl({1\over\Vert x\Vert}-{1\over \Vert y\Vert}\Bigr)=2.$$

So $X$ is strictly convex if and only if for distinct norm one vectors $x$, $y$, $\Vert x+y\Vert<2$.

$X$ is uniformly convex if and only if for each $\epsilon>0$, there is a $\delta>0$ so that for any $x$, $y$ in the closed unit ball of $X$ with $\Vert x-y\Vert\ge\epsilon$ one has $$\Bigl\Vert{ x+y\over 2}\Bigr\Vert\le 1-\delta.$$

It should be easy to see from the above that a uniformly convex normed space is strictly convex.

I think this property is not verified in uniformly convex Banach spaces in general, in fact it is not even true in the uniformly convex Banach space $(\mathbb{R},|.|)$: $|1+2|=|1|+|2|$