Bounded in Norm .... Garling, Section 11.2: Normed Spaces ....

[Math Amateur]

TL;DR Summary

The thread concerns the link between Garling's definition of a bounded set and the condition of a set being norm bounded or bounded in norm ...

I am reading D. J. H. Garling's book: "A Course in Mathematical Analysis: Volume II: Metric and Topological Spaces, Functions of a Vector Variable" ... ...

I am focused on Chapter 11: Metric Spaces and Normed Spaces ... ...

I need some help with some remarks by Garling concerning a subset being norm bounded of bounded in norm ...

The particular remarks by Garling read as follows:

Note that the definition of a bounded set by Garling is included in the following text:

In the remarks by Garling above we read the following:

" ... ... Then since

$\mid \mid x - y \mid \mid \le \mid \mid x \mid \mid + \mid \mid y \mid \mid$ and $\mid \mid y \mid \mid \le \mid \mid y - x \mid \mid + \mid \mid x \mid \mid$

a subset $B$ is bounded if and only if $\text{sup} \{ \mid \mid b \mid \mid \ : \ b \in B \} \lt \infty$ ... ... ... "Can someone please explain/demonstrate how (given Garling;s definition of a bounded subset) that the statements:

$\mid \mid x - y \mid \mid \le \mid \mid x \mid \mid + \mid \mid y \mid \mid$ and $\mid \mid y \mid \mid \le \mid \mid y - x \mid \mid + \mid \mid x \mid \mid$

lead to the statement that:

a subset $B$ is bounded if and only if $\text{sup} \{ \mid \mid b \mid \mid \ : \ b \in B \} \lt \infty$ ... ... ... ?Hope someone can help ...

Peter
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It may help some readers of the above post to have access to the start of Garling's section on normed spaces in order to familiarize them with Garling's approach and notation ... so I am providing the same ... as follows:

Hope that helps ...

Peter

 

[pasmith]

You have a normed vector space $E$ with norm $\|\cdot\|$. How is the metric defined in terms of this norm?

What happens if you substitute that into the definition of "bounded"?

There are two implications to prove to show that "bounded" is equivalent to "bounded in norm":

• $\|b\|$ bounded above implies $d(b,b')$ bounded above.
• $d(b,b')$ bounded above implies $\|b\|$ bounded above.

The remark gives two inequalities. What can you do with them?

 

[Math Amateur]

Hi pasmith ... thanks for your guidance and help ...

I received a proof (by Olinguito) on another website and the proof follows the lines suggested by you ...

Here is the suggested proof ...

$E$ is both a metric space and a vector space; the metric and the norm are related by
$d(x,y)\ =\ \|x-y\|$
for $x,y\in E$.

So if $B\subset E$ is nonempty – say it contains $c$ – and bounded, we have for all $b\in B$,
$\|b\|\ \le\ \|b-c\|+\|c\|\ \le\ \text{diam}(B)+\|c\|$
$\displaystyle\implies\ \sup_{b\in B}\|b\|<\infty$ (since $\text{diam}(B)<\infty$ and $\|c\|$ is a fixed number).

\Conversely, if $\displaystyle s=\sup_{b\in B}\|b\|<\infty$, then for all $b,b'\in B$,
$d(b,b')\ =\ \|b-b'\|\ \le\ \|b\|+\|b'\|\ \le\ 2s$
$\implies\ \text{diam}(B)<\infty$Peter