# Maximum of two norms is norm

How to prove triangle inequality for $$\|x\|=\max\left\{\|x\|_\alpha, \|x\|_\beta\right\}$$. I need to prove that $$\|\cdot\|$$ is a norm.

You need to prove the three properties of a norm.

First, $$\lvert\lvert{x}|| = 0$$ then $$x=0$$. That holds trivially because if the max of two positive numbers is zero, both numbers are zero, and because inside the max you have two norms.

Second, $$\lvert\lvert{ax}\rvert\rvert = |a| \lvert\lvert x\rvert\rvert$$ for $$a$$ real. that also holds trivially, since inside the max you have two norms.

Finally, you have to prove the triangle inequality, which is $$\lvert\lvert{x + y}\rvert\rvert \leq \lvert\lvert{x}\rvert\rvert + \lvert\lvert{y}\rvert\rvert$$.

To do that, write

\begin{align*} \lvert\lvert{x + y}\rvert\rvert &= \max\{\lvert\lvert{x + y}\rvert\rvert_a, \lvert\lvert{x + y}\rvert\rvert_b\}\\ &\leq \max\{\lvert\lvert{x }\rvert\rvert_a + \lvert\lvert{y }\rvert\rvert_a, \lvert\lvert{x}\rvert\rvert_b+ \lvert\lvert{y }\rvert\rvert_b\}\\ &\leq \max\{\lvert\lvert{x }\rvert\rvert_a, \lvert\lvert{x }\rvert\rvert_b\} + \max\{\lvert\lvert{y}\rvert\rvert_a, \lvert\lvert{y }\rvert\rvert_b\}\\ &=\lvert\lvert{x}\rvert\rvert + \lvert\lvert{y}\rvert\rvert \end{align*}

For the first inequality, we used the Triangle Inequality for both norms $$a$$ and $$b$$. For the second inequality, we used the property that the max of the sum is less than the sum of the max.

Hint: $$\|x+y\|_\alpha\le\|x\|_\alpha+\|y\|_\alpha\le \|x\|+\|y\|.$$ Do the same for $$\beta$$.