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I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

6.A.17 Prove or disprove: there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $(x,y)\in\mathbb{R}^2$.

I solved this exercise but I am worried if my solution is ok because this exercise appears to be unnaturally too easy.

My solution is the following:

If there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $(x,y)\in\mathbb{R}^2$, then $0<||(-1,-1)||=\max\{-1,-1\}=-1$.
This is a contradiction.

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    $\begingroup$ Yep, that's a good counterexample. $\endgroup$
    – Tom Chen
    May 10, 2022 at 2:41
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    $\begingroup$ This is correct, assuming no transcription errors from the question. A trickier (and IMHO more worthwhile) problem is to show that $\|(x, y)\| = \max \{|x|, |y|\}$ cannot be generated by an inner product (and it wouldn't surprise me if this is what Axler intended). $\endgroup$ May 10, 2022 at 2:41
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    $\begingroup$ I would bet serious money that @TheoBendit is right, and that it's a misprint. If you want to understand linear algebra, you should definitely be able to answer the "corrected" version. $\endgroup$
    – JonathanZ
    May 10, 2022 at 3:21
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    $\begingroup$ The missing absolute values in Exercises 17 and 18 in Section 6.A of Linear Algebra Done Right are typos that will be corrected in the next edition of the book. $\endgroup$ May 10, 2022 at 14:19
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    $\begingroup$ @tchappy ha I am working on the fourth edition of Linear Algebra Done Right now. It is still at least a year away from completion. $\endgroup$ May 11, 2022 at 19:15

2 Answers 2

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As covered in the comments, you need absolute values to get what is called the $\infty$-norm: $$\|(x, y)\|_{\infty} = \max\{|x|, |y|\}.$$ This is the limit as $p \to \infty$ of the $p$-norm $$\|(x, y)\|_p = (|x|^p + |y|^p)^{1/p}$$ which generalizes the Euclidean norm ($p = 2$).

The rule for when a norm comes from an inner product is called the Parallelogram law which says that if $\| \cdot \|$ comes from an inner product, then for all $u, v$ $$\|u + v\|^2 + \|u - v\|^2 = 2(\|u\|^2 + \|v\|^2).$$ Moreover, if this identity holds for all $u$ and $v$, the inner product is recovered "by polarization" as $$\langle u, v \rangle = \frac{\|u + v\|^2 - \|u - v\|^2}{4}.$$

So continuing with Axler's exercise, show that the $\infty$-norm (with the absolute values) does not satisfy the Parallelogram law. Slightly more challenging: show that the Parallelogram law holds for all $u, v$ for the $p$-norm if and only $p = 2$. I recommend thinking about an actual parallelogram.

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  • $\begingroup$ Thank you very much for your perfect answer. $\endgroup$
    – tchappy ha
    May 10, 2022 at 4:30
  • $\begingroup$ By chance, the next exercise (6.A.18) is "Suppose $p>0$. Prove that there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $||(x,y)||=(x^p+y^p)^{1/p}$ for all $(x,y)\in\mathbb{R}^2$ if and only if $p=2$." Are you a prophet? $\endgroup$
    – tchappy ha
    May 10, 2022 at 4:40
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    $\begingroup$ @tchappyha It's only a norm if $p \ge 1$ (otherwise it doesn't satisfy the triangle inequality). Also you need absolute values again. This family of norms (including $p = \infty$) is very common and if you're discussing which norms come from an inner product, you're all but required to include such an exercise or example. $\endgroup$ May 10, 2022 at 4:53
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Thank you very much Theo Bendit.

Prove or disprove: there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{|x|,|y|\}$$ for all $(x,y)\in\mathbb{R}^2$.

My solution:

Assume that $\langle(x,y),(x,y)\rangle=\max\{|x|,|y|\}^2$.
Then, $\langle(1,0),(1,0)\rangle=\langle(0,1),(0,1)\rangle=1$.
Then, $1=\langle(1,1),(1,1)\rangle=\langle (1,0)+(0,1),(1,0)+(0,1)\rangle=\langle(1,0),(1,0)\rangle+2\langle(1,0),(0,1)\rangle +\langle(0,1),(0,1)\rangle=1+2\langle(1,0),(0,1)\rangle +1=2+2\langle(1,0),(0,1)\rangle$.
So, $\langle(1,0),(0,1)\rangle=-\frac{1}{2}$.
Then, $2=\langle(1,2),(1,2)\rangle=\langle (1,0)+(0,2),(1,0)+(0,2)\rangle=\langle(1,0),(1,0)\rangle+4\langle(1,0),(0,1)\rangle +4\langle(0,1),(0,1)\rangle=1-2+4=3$.
This is a contradiction.

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