My Question is:
Sketch the unit ball $B(0, 1)$ in $\mathbb{R}^2$ equipped with the following norm: $||(x, y)|| =$ max{|$x$|,|$y$|}
I'm semi confident in this topic but cant seem to find the right graph to sketch so any help will be appreciated.
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By definition, $B(0,1)$ is the set of all points $(x,y)$ satisfying $||(x,y)|| < 1$. Does that help? – user49640 Feb 20 '17 at 21:21
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Simply recall what the unit ball is, it is the set of $\mathbf{x}\in\mathbb{R}^2$ such that $\|\mathbf{x}\|\leqslant 1$. If we set, $\mathbf{x}:=(x,y)$, one has: $$\mathbf{x}\in B(0,1)\Leftrightarrow |x|\leqslant 1\textrm{ and }|y|\leqslant 1.$$ Hence $B(0,1)$ is the square center at $(0,0)$ of length $2$ and whose sides are respectively parallel to the $x$-axis and the $y$-axis. A drawing is probably clearer:
In the same fashion, can you sketch the unit ball of $\mathbb{R}^2$ endowed with the following norm: $$\|(x,y)\|_1:=|x|+|y|?$$
C. Falcon
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Thank you for the sketch it makes a lot of sense now. Im not sure about sketching $|(x,y)|_1:=|x|+|y|$ could you try and explain this one? – AFraggers Feb 20 '17 at 21:54
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We are trying to solve $|x|+|y|\leqslant 1$, it is a good idea to distinguish cases, namely $x$ and $y$ positive, $x$ and $y$ negative, $x$ positive and $y$ negative, $x$ negative and $y$ positive. In the first case, the inequality is $x+y\leqslant 1$, so that we get all the area in the first quadrant below the line $y=1-x$. – C. Falcon Feb 20 '17 at 22:09