> [!NOTE] > In any [[Norm|normed space]] $(X,\|\cdot\|)$, the closed unit ball $\boldsymbol{B}_X$ is [[Convexity|convex]] and conversely suppose that a function $\|\cdot\|: X \rightarrow \mathbb{R}^{+}$satisfies (i) and (ii) of the definition of $a$ norm and let $\mathfrak{B}:=\{x \in X:\|x\| \leq 1\}$. Then $\|$ - $\|$ satixfies the triangle incquality (and hence is a norm) if and only if $\mathfrak{B}$ is conver. ###### Proof: If $x, y \in \mathfrak{B}_X$ then $\|x\| \leq 1$ and $\|y\| \leq 1$. Then, for $0<\lambda<1$, $ \|\lambda x+(1-\lambda) y\| \leq|\lambda|\|x\|+|1-\lambda|\|y\| \leq \lambda+(1-\lambda)=1, $ so $\lambda x+(1-\lambda) y \in \boldsymbol{B}_x$. We now prove the converse direction. Suppose $\Vert x\Vert, \Vert y \Vert >0$ since if $\|x\|=0$ then $x=0$ and $\|x+y\|=\|y\|=\|x\|+\|y\|$, so the triangle inequality holds. In this case, both $x /\|x\|, y /\|y\| \in \mathfrak{B}$, so using the convexity of $\mathfrak{B}$ we have $ \frac{\|x\|}{\|x\|+\|y\|}\left(\frac{x}{\|x\|}\right)+\frac{\|y\|}{\|x\|+\|y\|}\left(\frac{y}{\|y\|}\right) \in \mathfrak{B} $ which can be simplified into $ \frac{x+y}{\|x\|+\|y\|} \in \mathfrak{B} $ Using property (ii) from Definition 2.1 it follows that $ \left\|\frac{x+y}{\|x\|+\|y\|}\right\|=\frac{\|x+y\|}{\|x\|+\|y\|} \leq 1 \Rightarrow\|x+y\| \leq\|x\|+\|y\|, $ as required.