Minimum of two norms in $\mathbb R^2$ needs not be a norm I know that if $\|x\|$ and $\|x\|'$ are norms, then $\min (\|x\|,\|x\|')$  needs not be a norm. 
Is there any counterexample in $\mathbb R^2$ which shows that $\min (\|x\|,\|x\|')$ really is not a norm? This should probably be easy but I can't find one.
 A: The canonical norms on $\mathbb R^2$ are 
$$
\|(x,y)\|_\infty=\max\{|x|,|y|\},\ \ \|(x,y)\|_1=|x|+|y|,\ \ \|(x,y)\|_2=(x^2+y^2)^{1/2}.
$$
An example cannot come directly from these, because we have $\|\cdot\|_\infty\leq\|\cdot\|_2\leq\|\cdot\|_1$. But let us construct a new norm,
$$
\|(x,y)\|_c=\frac{2\|(x,y)\|_\infty+\|(x,y)\|_1}3.
$$
Now, if $$\alpha(x,y)=\min\{\|(x,y)\|_c,\|(x,y)\|_2\},$$ we have 
$$
\left(2,3/2\right)=\left(1,1/2\right)+(1,1).
$$
We also have 
$$
\alpha(2,3/2)=\min\left\{\frac{4+2+3/2}3,\sqrt{4+\frac94}\right\}=\min\left\{\frac52,\frac52\right\}=\frac52,
$$
$$
\alpha(1,1/2)=\min\left\{\frac{2+3/2}3 ,\frac{\sqrt5}2\right\}=\frac{\sqrt5}2,
$$
$$
\alpha(1,1)=\min\left\{\frac{2+1+1}3,\sqrt2 \right\}=\frac43.
$$
As $\sqrt5/2+4/3<2.46<5/2$, 
$$
\alpha((1,1/2)+(1,1))> \alpha(1,1/2)+\alpha(1,1),
$$
contradicting the triangle inequality. So $\alpha$ is not a norm. 
A: Let $\|(x,y)\|_1=|x|+2|y|$ and $\|(x,y)\|_2=2|x|+|y|.$ 
Let $f(p)=\min(\|p\|_1,\|p\|_2).$ 
Let $p_1=(2,1)$ and $p_2=(1,2).$ 
Then $f(p_1)+f(p_2)=4+4=8<9=f(p_1+p_2).$
