Let $\Vert x\Vert = \max\Big\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Big\}$ with $x \in \mathbb{R}^2$. Show that $\Vert\cdot\Vert$ is a norm on $\mathbb{R}^2$.

I do have problems showing that the triangle inequality holds.

So far, I tried the following: Let $x, y \in \mathbb{R}^2$, than

$$ \begin{align*} \Vert x\Vert &= \max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\}\\ &= \frac{1}{2}\left(\frac{|x_1|}{3} + \frac{|x_2|}{2} + \Bigg|\frac{|x_1|}{3} - \frac{|x_2|}{2}\Bigg|\right)\\ &=\frac{1}{6}\left( \frac{1}{2}\left(2|x_1| + 3|x_2| + \Big|2|x_1| -3|x_2|\Big|\right)\right)\\ &=\frac{1}{6}\max\Big\{2|x_1|, 3|x_2|\Big\}, \end{align*} $$


$$\small \begin{align*} \Vert x + y \Vert &\leq \Vert x \Vert + \Vert y \Vert\\ \Leftrightarrow \max\Bigg\{\frac{|x_1 + y_1|}{3}, \frac{|x_2 + y_2|}{2}\Bigg\} &\leq \max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\} + \max\Bigg\{\frac{|y_1|}{3}, \frac{|y_2|}{2}\Bigg\}\\ \Leftrightarrow \frac{1}{6}\max\Big\{2|x_1 + y_1|, 3|x_2+y_2|\Big\} &\leq \frac{1}{6}\left(\max\Big\{2|x_1|, 3|x_2|\Big\} + \max\Big\{2|y_1|, 3|y_2|\Big\}\right)\\ \Leftrightarrow 2|x_1+y_1| + 3|x_2+y_2| + \Big|2|x_1+y_1| -3|x_2+y_2|\Big| &\leq 2|x_1| + 3|x_2| + \Big|2|x_1| -3|x_2|\Big| + 2|y_1| + 3|y_2| + \Big|2|y_1| -3|y_2|\Big|\\ \Leftrightarrow \underbrace{2|x_1+y_1| + 3|x_2+y_2|}_{\leq\left(2|x_1| +2|x_2|\right) + \left(3|x_1| + 3|x_2|\right)} + \Big|2|x_1+y_1| -3|x_2+y_2|\Big| &\leq \left(2|x_1| +2|x_2|\right) + \left(3|x_1| + 3|x_2|\right) + \Big(\Big|2|x_1| -3|x_2|\Big| + \Big|2|y_1| -3|y_2|\Big|\Big) \end{align*} $$

I have problems showing the last inequality. I do know that $\Big|2|x_1+y_1| -3|x_2+y_2|\Big| \leq \Big|2|x_1| -3|x_2|\Big| + \Big|2|y_1| -3|y_2|\Big|$ is not true in general. So I suppose there is something I do not see?

  • $\begingroup$ An alternative solution is to notice that this is the well-known $\ell^\infty = \|-\|_\infty$ norm on $T(\mathbb{R}^2)$, where $T$ is the diagonal matrix with $t_{11} = 1/3$, $t_{22} = 1/2$. This greatly simplifies the proof: $$\|T(x+y)\|_\infty = \|Tx + Ty\|_\infty \leqslant \|Tx\|_\infty + \|Ty\|_\infty$$ by the fact that we know $\|-\|_\infty$ is a norm. $\endgroup$
    – William
    May 10, 2021 at 3:47

1 Answer 1


To prove that the triangles inequality holds, we need to show that $$\|x+y\|\leq \|x\|+\|x\|$$ $$\Leftrightarrow \max\Bigg\{\frac{|x_1 + y_1|}{3}, \frac{|x_2 + y_2|}{2}\Bigg\} \leq \max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\} + \max\Bigg\{\frac{|y_1|}{3}, \frac{|y_2|}{2}\Bigg\} $$ which is equivalent to $$\frac{|x_1 + y_1|}{3} \leq \max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\} + \max\Bigg\{\frac{|y_1|}{3}, \frac{|y_2|}{2}\Bigg\} $$ and $$\frac{|x_2 + y_2|}{2} \leq \max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\} + \max\Bigg\{\frac{|y_1|}{3}, \frac{|y_2|}{2}\Bigg\} $$ Note that $$\frac{|x_1 + y_1|}{3} \leq \frac{|x_1|}{3}+\frac{|y_1|}{3}\leq\max\Bigg\{\frac{|x_1|}{3}, \frac{|x_2|}{2}\Bigg\} + \max\Bigg\{\frac{|y_1|}{3}, \frac{|y_2|}{2}\Bigg\}$$ The second inequality can be obtained similarly.

  • 1
    $\begingroup$ +1 The OP should look t this, see how much clearer it is than his attempt, and learn from it. $\endgroup$
    – GEdgar
    Apr 4, 2021 at 11:01
  • $\begingroup$ @GEdgar thank you! $\endgroup$
    – JwJJJJ
    Apr 5, 2021 at 1:20

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