Show that a subset of a normed space is bounded iff its image by any operator is bounded

Question

Could you give me a hint with the following exercise?

Let $X$ be a $\mathbb{K}$-normed vector space and $Y\subset X$, show that $Y$ is bounded respect to the norm on $X$, iff, for all $T\in X'$, $T(Y)$ is bounded on $\mathbb{K}$.

Thanks.

Edited

Using Kavi's hint:

[$\Rightarrow$] Let $T\in X'$ be arbitrary, then $T$ is continuous, as $Y$ is bounded, then $T(Y)$ is bounded.

[$\Leftarrow$] Suppose $T(Y)$ is bounded for every $T \in X'$. If possible let $Y$ be unbounded so that $\|y_n\| \to \infty$ for some sequence $(y_n)$ in $Y$. Define $S_n: X' \to K$ by $S_n(T)=T(y_n)$. So, $$\|S_n(T)\|=\|T(y_n)\| $$ But, $\{\|T(y_n)\|\}$ is bounded for all $T\in X'$ by hyphotesis, so $\{\|S_n(T)\|\}$ is bounded for all $T\in X'$.

Applying Uniform Boundedness Principle (alias Banach–Steinhaus theorem) we get that $$\{ \|S_n\| \}$$ is bounded, so there is $C>0$ such that $$\|S_n\|\leq C,\forall n\in\mathbb{N}\quad\Rightarrow \sup\{\|S_n(T)\|:\|T\|=1\}=\|S_n\|\leq C,\forall n\in\mathbb{N} $$

$$\Rightarrow \|T(y_n)\|\leq C, \|T\|=1, \forall n\in\mathbb{N}.$$

On the other hand, by the Hanh-Banach Theorem, for every $x\in X$, $$\|x\|=\sup\{|T(x)|,\|T\|=1\}$$ in particular for $y_n\in X$ $$\|y_n\|=\sup\{|T(y_n)|,\|T\|=1\}\leq C$$ is a contradiction.

Kavi Rama Murthy · Answer 1 · 2020-02-14T08:08:26.890

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Hint: Suppose $T(Y)$ is bounded for every $T \in X'$. If possible let $Y$ be unbounded so that $\|y_n\| \to \infty$ for some sequence $(y_n)$ in $Y$. Define $S_n: X' \to K$ by $S_n(T)=T(y_n)$. Apply Uniform Boundedness Principle (alias Banach–Steinhaus theorem) to get a contradiction. The reverse implication is obvious.

[Uniform Boundedness Principle gives you $|Ty_n| \leq C $ for all $n$ and for all $T$ with $\|T\| \leq 1$. If you fix $n$ there exists $T$ such that $\|T\|=1$ and $Ty_n=\|y_n\|$. [This is a standard application of Hahn Banach Theorem]. Hence we get $\|y_n\| \leq C$ for all $n$].

edited Feb 14 '20 at 08:08

answered Feb 13 '20 at 06:40

Kavi Rama Murthy

311,013

The UniformBoundedness Principle applies to Banach spaces but not to all normed linear spaces. E.g. let $X \subset l_{\infty}$ where $(x_n)_n\in X$ iff ${n: x_n\ne 0}$ is finite. For $j\in \Bbb N$ let $f_j((x_n)_n)=jx_j.$ For any $x\in X$ the set ${j: f_j(x)\ne 0}$ is finite, so ${f_j}_j$ is point-wise bounded on $X$. But $|f_j|=j.$ – DanielWainfleet Feb 13 '20 at 10:22
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@D I am applying UBP to operators from $X'$ to $K$. $X'$ is complete even if $X$ is not. – Kavi Rama Murthy Feb 13 '20 at 10:30
Thank you very much to both of you, i will try with that idea. – Framate Feb 13 '20 at 18:04
@KaviRamaMurthy i edited trying to solve, can you take a look please. Thank you – Framate Feb 14 '20 at 07:59
@Framate I have added some details. Hover the mouse over the yellow portion to see the details. – Kavi Rama Murthy Feb 14 '20 at 08:09
@Framate What we need is an operator from $X$ to the scalars. So the identity map if $Y$ is not useful here. Hahn Banach Theorem is necessary. – Kavi Rama Murthy Feb 14 '20 at 08:41
I see, i will think better – Framate Feb 14 '20 at 08:46
@KaviRamaMurthy I finished, thank you very much!! – Framate Feb 14 '20 at 09:22

Show that a subset of a normed space is bounded iff its image by any operator is bounded

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