TITLE: Why a function satifies certain conditions is a norm?
QUESTION [1 upvotes]: Let $E$ a linear normed space over $\mathbb{R}$ and $n:E\to \mathbb{R}_+$ an application such that: $n(x)=0\Leftrightarrow x=0$, $n(\alpha x)= |\alpha|n(x)$ and the set $$A=\{ x\in E:n(x)\leq 1\}$$ is convex. Then we have to prove that $n$ defines a norm. 
My approach: Is sufficient to prove that satisfies the triangle inequality. For $x$ and $y$ in $A$ with norm equal to 1 i have prove the result. My problem is to generalize!!!
Thanks in advance!!

REPLY [1 votes]: Let $x,y\in E$. If $x=0$ por $y=0$ then it is clear that $n(x+y)\leq n(x)+n(y)$, since $n(0)=0$, so we can assume that $x\neq 0$ and $y\neq 0$.
Now, le $x'=x/n(x)$ and $y'=y/n(y)$. Note that $n(x')=n(y')=1$, so $x',y'\in A$. For every $0\leq t\leq 1$, $tx'+(1-t)y'\in A$ since $A$ is convex. This means that
$n(tx'+(1-t)y')\leq 1$ and hence, 
$$
n\left(\frac{tx}{n(x)}+\frac{(1-t)y}{n(y)}\right)\leq 1
$$.
Choose $t=\frac{n(x)}{n(y)+n(x)}$. Note that $0<t<1$ and that $1-t=\frac{n(y)}{n(y)+n(x)}$, so we can substitute this value in the last equation to obtain
$$
n\left(\frac{x}{n(x)+n(y)}+\frac{y}{n(x)+n(y)}\right)\leq 1
$$
The left hand side is equal to $\frac{n(x+y)}{n(x)+n(y)}$ so this inequality is equivalent to the triangle inequality for $n$.
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