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CONTEXT

I'm writing a proof. I'm trying to make sure my mathematical writing is correct.

I want verify that the supremum $$\sup\left\{\left\|\textbf{x}\right\|_p \mid \textbf{x} \in B_r{\left(\bf{0}\right)}\right\} = r$$

QUESTION.

Can you verify my proof?

PROOF.

From [A], the $p$-norm of vector $ \mathbf {x} =(x_{1},\ldots ,x_{m}) $ is $$\left\| \mathbf{x} \right\| _p := \bigg( \sum_{i=1}^m \left| x_i \right| ^p \bigg) ^{1/p}.$$

From [B], open ball of radius $r$ around the origin ${\bf 0}$ is the subset $$B_r{\left({\bf 0 }\right)} = \left\{{\bf y}\in \mathbb{R}^m \mid \left\|{\bf 0 } - {\bf y }\right\|_2<r\right\}. $$

Let the set $S$ be given as $$S =\{ \left\| \mathbf{y} \right\| _p \mid {\bf{y}} \in B_r{\left(\bf{0}\right)}, p \geq 1~\textrm{(real)},~m>0~\textrm{(integer)} \}$$

As a consequence, paraphrasing liberally from the formal definition of upper bound and supremum in [C], I write, ``... an upper bound of the subset $S$ of a partially ordered set $(\mathbb{R},\leq)$ is an element $r$ of $\mathbb{R}$ such that $$ r ≥ \left\| \mathbf{x} \right\| _p ~\textrm{for all}~\left\| \mathbf{x}\right\|_p \in S.$$ The upper bound $r$ of $S$ is called a supremum of $S$ since for all upper bounds $z$ of $S$ in $\mathbb{R}$, $z \geq r$...''

I conclude that the supremum $$\sup\left\{\left\|\textbf{x}\right\|_p \mid \textbf{x} \in B_r{\left(\bf{0}\right)}\right\} = r$$

BIBLIOGRAPHY

[A] https://en.wikipedia.org/wiki/Norm_(mathematics)

[B] Hubbard and Hubbard, Vector calculus, linear algebra, and differential forms: a unified approach, 2nd ed., pg 90.

[C] https://en.wikipedia.org/wiki/Infimum_and_supremum#Formal_definition

Michael Levy
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    Well, since no matter the $x\in\Bbb R^n$, there will always be a $y$ with bigger norm (like $y=2x$), your definition seems to make that max-y thing nonexistent. Are you sure that you meant what you wrote? – Lubin Apr 12 '19 at 18:08
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    I think that the more conventional thing would be to put “$x\in B(r)$” beneath the “max”, instead of just “$x$”. – Lubin Apr 13 '19 at 15:40
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    It's perhaps easier to visualise in $p=1=n$. The set $[0,r)$ has no maximum, but it does have a supremum. – Calvin Khor Apr 16 '19 at 13:19

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