Prove that this is a norm

Question

Let $X$ be a real or complex vector space. Let $N:X\rightarrow\mathbb{R}$ be a function with the following properties

1. $N(x)\geq 0$,
2. $N(x)=0$ if and only if $x=0$,
3. $N(\lambda x)=|\lambda|N(x)$,
4. The set $B=\lbrace x\in X\text{ }|\text{ }N(x)\leq 1\rbrace$ is convex.

Prove that $N$ is a norm.

I am unable to prove the triangle inequality.

Answer 1

Consider $x,y\ne 0$ and define $\tilde{x}=\frac{1}{N(x)}x$, $\tilde{y}=\frac{1}{N(y)}y$. Then, by property 3., $N(\tilde{x})=1=N(\tilde{y})$. So $\tilde{x},\tilde{y}$ belong to $B$. Let's use convexity then: $$ z=\frac{N(x)}{N(x)+N(y)}\tilde{x} + \frac{N(y)}{N(x)+N(y)}\tilde{y} \in B.$$ This yields $N(z)\leq 1$ which is equivalent to $$N(N(x)\tilde{x}+N(y)\tilde{y})\leq N(x)+N(y)$$by property 3. again.

Now, since $N(x)\tilde{x}=x$ and $N(y)\tilde{y}$, by definition, we're done.

Answer 2

Hint: Show $$\frac{x+y}{N(x)+N(y)}\in B$$