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

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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.

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

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