Visualising a region involving the $1$-norm I'm a long-time lurker but first-time poster so I apologise if this post isn't formatted correctly. I'm having trouble visualising the region $\mathcal{R}$ defined as: $$\mathcal{R}=\\{\textbf{x}\in\mathbb{R}^2:\Vert{\textbf{x}}\Vert_1\leq\lambda, \lambda>0\\}$$ I know that the unit ball for the $1$-norm is a diamond shape, so I thought that $\mathcal{R}$ would be the diamond in the first quadrant (i.e. a triangle with the vertices $(0,0),\ (1,0),\ (0,1)).$ But my lecturer said that $\mathcal{R}$ is a square with the vertices $(0,0),\ (0,1),\ (1,0),\ (1,1).$ I know this problem is trivial but there is obviously a hole in my understanding which I'm hoping you can help me fix. Thanks in advance.

It is a square. Let inspect the unit norm ball. $$\mathcal{R}=\\{\textbf{x}\in\mathbb{R}^2:\Vert{\textbf{x}}\Vert_1\leq 1\\}$$

Let say, $\Vert{\textbf{x}}\Vert_1= 1.$ So, $$|x_1|+|x_2|=1.$$ **Case-1** When, $x_1,\ x_2 \geq 0,$ $$x_2=1-x_1.$$

**Case-2** When, $x_1 > 0,\ x_2 < 0,$ $$x_2=x_1-1.$$

**Case-3** When, $x_2 > 0,\ x_1 < 0,$ $$x_2=x_1+1.$$

**Case-4** When, $x_1, \ x_2 < 0,$ $$x_2=-(x_1+1).$$

So, now you can draw the region, and you will find that it is a square having vertices $(-1,0),\ (1,0), \ (0,1),\ (0,-1).$