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Weak-* continuous functionals with bounded level sets

Weakest topology with respect to which ALL linear functionals are continuousTopological modules with enough continuous linear functionals.Does every continuous operator between normed spaces map bounded sets to bounded sets?Representation of weak*-continuous linear functionals on a subspace which is not necessarily weak* closedSet of poinwise convergence of a sequence of weak* continuous linear functionals is weak* closed?Separation of a weak-star closed wedge by a weak-star continuous linear functionalWeak-* Convergence of functionals defined by probability measures ${delta_n}$Function whose range is a bounded subset of an infinite-dimensional Banach space is weakly continuous?are linear functionals on C[0, 1] bounded and thus continuousHahn-Banach in dual space

1

1

$begingroup$

Let $X$ be an infinite-dimensional Banach space, $X^*$ its topological dual and $f:X^*tomathbb{R}$ some weak-* continuous functional (not necessarily linear).

Is it possible for $f^{-1}(a)$ to be nonempty and bounded?

Attempt: I thought perhaps the operator norm $|cdot|_{op}$ on $X^*$ would be an example, but it is actually not even weak-* continuous since weak-* open sets in infinite dimensions are not bounded. Linear functionals clearly don't work either.

---

Follow-up: If the weak-* continuous functionals cannot have nonempty bounded level sets, then a natural follow-up question would be:

Is it possible for $f^{-1}(a)$ to have a connected component which is unbounded?

functional-analysis topological-vector-spaces weak-topology

share|cite|improve this question

edited Mar 13 at 19:37

Christian Bueno

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want to have this for one $a$ or for all $a$?
  $endgroup$
  – gerw
  Mar 14 at 19:35
  

  
  
  
  

add a comment | 

1

1

$begingroup$

Let $X$ be an infinite-dimensional Banach space, $X^*$ its topological dual and $f:X^*tomathbb{R}$ some weak-* continuous functional (not necessarily linear).

Is it possible for $f^{-1}(a)$ to be nonempty and bounded?

Attempt: I thought perhaps the operator norm $|cdot|_{op}$ on $X^*$ would be an example, but it is actually not even weak-* continuous since weak-* open sets in infinite dimensions are not bounded. Linear functionals clearly don't work either.

---

Follow-up: If the weak-* continuous functionals cannot have nonempty bounded level sets, then a natural follow-up question would be:

Is it possible for $f^{-1}(a)$ to have a connected component which is unbounded?

functional-analysis topological-vector-spaces weak-topology

share|cite|improve this question

edited Mar 13 at 19:37

Christian Bueno

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want to have this for one $a$ or for all $a$?
  $endgroup$
  – gerw
  Mar 14 at 19:35
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ be an infinite-dimensional Banach space, $X^*$ its topological dual and $f:X^*tomathbb{R}$ some weak-* continuous functional (not necessarily linear).

Is it possible for $f^{-1}(a)$ to be nonempty and bounded?

Attempt: I thought perhaps the operator norm $|cdot|_{op}$ on $X^*$ would be an example, but it is actually not even weak-* continuous since weak-* open sets in infinite dimensions are not bounded. Linear functionals clearly don't work either.

---

Follow-up: If the weak-* continuous functionals cannot have nonempty bounded level sets, then a natural follow-up question would be:

Is it possible for $f^{-1}(a)$ to have a connected component which is unbounded?

functional-analysis topological-vector-spaces weak-topology

share|cite|improve this question

edited Mar 13 at 19:37

Christian Bueno

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

$endgroup$

share|cite|improve this question

edited Mar 13 at 19:37

Christian Bueno

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

share|cite|improve this question

edited Mar 13 at 19:37

Christian Bueno

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

asked Mar 12 at 22:05

Christian BuenoChristian Bueno

632415

1 Answer
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active

oldest

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0

$begingroup$

Yes, there is such a function. Let $X=ell^2$ with its standard norm, and identify $X^*$ with $X$ via the mapping $ymapsto langle -,yrangle$. Denote the standard orthonormal basis by $e_n$. Define $f:X^*tomathbb C$ by
$$f(y)=sum_{ninmathbb N} |langle 2^{-n}e_n,yrangle|.$$

First, that this sum is convergent for any fixed $y$: $|langle e_n,yrangle|le|y|$, so the sum of nonnegative terms is bounded above by $sum_{ninmathbb N} 2^{-n}|y|<infty$.

If $f$ were weak-star continuous, it would provide an example of such a function. $f(0)=0$, so the preimage of $0$ is nonempty. Assume that $f(y)=0$. Then $|langle e_n,yrangle|=0$ for each $e_n$, so $langle e_n,yrangle=0$, so $y$ is $0$ on a (Schauder) basis. So $y$ must be the zero functional, implying that $f^{-1}(0)={0}$, a bounded set.

Now we need only show the weak-star continuity. By separability of $ell^2$, it is enough to check the convergence of limits. Assume that ${y_j}_{jinmathbb N}xrightarrow{w*} y$. We need to show that $f(y_j)rightarrow f(y)$. Note that by the Banach-Steinhaus theorem, weak-star convergent sequences must be bounded. Pick $R$ such that $|y_j|<R$ for each $j$.

Now let $epsilon>0$. Pick $N$ large so that $sum_{n>N} frac{1}{2^n}R<frac{epsilon}{4}$. Pick $K_1,ldots,K_N$ large (from the weak-star convergence of ${y_j}$) so that
$$forall j>K_i,quad |langle e_i,(y-y_j)rangle|<frac{epsilon}{4N}$$
and let $M=text{max}{N,K_1,ldots,K_N}$. Then for all $j>M$,
begin{align*}
|f(y_j)-f(y)|&=bigg|sum_{ninmathbb N}|langle 2^{-n}e_n,y_jrangle|-sum_{ninmathbb N}|langle 2^{-n}e_n,yrangle|bigg|\
&lebigg|sum_{nle N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|+bigg|sum_{n>N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|\
&le bigg|sum_{nle N}frac{epsilon}{2N}bigg|+bigg|sum_{n>N} 2^{-n}cdot 2Rbigg|\
&< frac{epsilon}{2}+frac{epsilon}{2}\
&<epsilon
end{align*}
so the function is weak-star continuous.

share|cite|improve this answer

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

$endgroup$

add a comment | 

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0

$begingroup$

Yes, there is such a function. Let $X=ell^2$ with its standard norm, and identify $X^*$ with $X$ via the mapping $ymapsto langle -,yrangle$. Denote the standard orthonormal basis by $e_n$. Define $f:X^*tomathbb C$ by
$$f(y)=sum_{ninmathbb N} |langle 2^{-n}e_n,yrangle|.$$

First, that this sum is convergent for any fixed $y$: $|langle e_n,yrangle|le|y|$, so the sum of nonnegative terms is bounded above by $sum_{ninmathbb N} 2^{-n}|y|<infty$.

If $f$ were weak-star continuous, it would provide an example of such a function. $f(0)=0$, so the preimage of $0$ is nonempty. Assume that $f(y)=0$. Then $|langle e_n,yrangle|=0$ for each $e_n$, so $langle e_n,yrangle=0$, so $y$ is $0$ on a (Schauder) basis. So $y$ must be the zero functional, implying that $f^{-1}(0)={0}$, a bounded set.

Now we need only show the weak-star continuity. By separability of $ell^2$, it is enough to check the convergence of limits. Assume that ${y_j}_{jinmathbb N}xrightarrow{w*} y$. We need to show that $f(y_j)rightarrow f(y)$. Note that by the Banach-Steinhaus theorem, weak-star convergent sequences must be bounded. Pick $R$ such that $|y_j|<R$ for each $j$.

Now let $epsilon>0$. Pick $N$ large so that $sum_{n>N} frac{1}{2^n}R<frac{epsilon}{4}$. Pick $K_1,ldots,K_N$ large (from the weak-star convergence of ${y_j}$) so that
$$forall j>K_i,quad |langle e_i,(y-y_j)rangle|<frac{epsilon}{4N}$$
and let $M=text{max}{N,K_1,ldots,K_N}$. Then for all $j>M$,
begin{align*}
|f(y_j)-f(y)|&=bigg|sum_{ninmathbb N}|langle 2^{-n}e_n,y_jrangle|-sum_{ninmathbb N}|langle 2^{-n}e_n,yrangle|bigg|\
&lebigg|sum_{nle N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|+bigg|sum_{n>N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|\
&le bigg|sum_{nle N}frac{epsilon}{2N}bigg|+bigg|sum_{n>N} 2^{-n}cdot 2Rbigg|\
&< frac{epsilon}{2}+frac{epsilon}{2}\
&<epsilon
end{align*}
so the function is weak-star continuous.

share|cite|improve this answer

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

$endgroup$

add a comment | 

0

$begingroup$

Yes, there is such a function. Let $X=ell^2$ with its standard norm, and identify $X^*$ with $X$ via the mapping $ymapsto langle -,yrangle$. Denote the standard orthonormal basis by $e_n$. Define $f:X^*tomathbb C$ by
$$f(y)=sum_{ninmathbb N} |langle 2^{-n}e_n,yrangle|.$$

First, that this sum is convergent for any fixed $y$: $|langle e_n,yrangle|le|y|$, so the sum of nonnegative terms is bounded above by $sum_{ninmathbb N} 2^{-n}|y|<infty$.

If $f$ were weak-star continuous, it would provide an example of such a function. $f(0)=0$, so the preimage of $0$ is nonempty. Assume that $f(y)=0$. Then $|langle e_n,yrangle|=0$ for each $e_n$, so $langle e_n,yrangle=0$, so $y$ is $0$ on a (Schauder) basis. So $y$ must be the zero functional, implying that $f^{-1}(0)={0}$, a bounded set.

Now we need only show the weak-star continuity. By separability of $ell^2$, it is enough to check the convergence of limits. Assume that ${y_j}_{jinmathbb N}xrightarrow{w*} y$. We need to show that $f(y_j)rightarrow f(y)$. Note that by the Banach-Steinhaus theorem, weak-star convergent sequences must be bounded. Pick $R$ such that $|y_j|<R$ for each $j$.

Now let $epsilon>0$. Pick $N$ large so that $sum_{n>N} frac{1}{2^n}R<frac{epsilon}{4}$. Pick $K_1,ldots,K_N$ large (from the weak-star convergence of ${y_j}$) so that
$$forall j>K_i,quad |langle e_i,(y-y_j)rangle|<frac{epsilon}{4N}$$
and let $M=text{max}{N,K_1,ldots,K_N}$. Then for all $j>M$,
begin{align*}
|f(y_j)-f(y)|&=bigg|sum_{ninmathbb N}|langle 2^{-n}e_n,y_jrangle|-sum_{ninmathbb N}|langle 2^{-n}e_n,yrangle|bigg|\
&lebigg|sum_{nle N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|+bigg|sum_{n>N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|\
&le bigg|sum_{nle N}frac{epsilon}{2N}bigg|+bigg|sum_{n>N} 2^{-n}cdot 2Rbigg|\
&< frac{epsilon}{2}+frac{epsilon}{2}\
&<epsilon
end{align*}
so the function is weak-star continuous.

share|cite|improve this answer

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

$endgroup$

add a comment | 

$begingroup$

Yes, there is such a function. Let $X=ell^2$ with its standard norm, and identify $X^*$ with $X$ via the mapping $ymapsto langle -,yrangle$. Denote the standard orthonormal basis by $e_n$. Define $f:X^*tomathbb C$ by
$$f(y)=sum_{ninmathbb N} |langle 2^{-n}e_n,yrangle|.$$

First, that this sum is convergent for any fixed $y$: $|langle e_n,yrangle|le|y|$, so the sum of nonnegative terms is bounded above by $sum_{ninmathbb N} 2^{-n}|y|<infty$.

If $f$ were weak-star continuous, it would provide an example of such a function. $f(0)=0$, so the preimage of $0$ is nonempty. Assume that $f(y)=0$. Then $|langle e_n,yrangle|=0$ for each $e_n$, so $langle e_n,yrangle=0$, so $y$ is $0$ on a (Schauder) basis. So $y$ must be the zero functional, implying that $f^{-1}(0)={0}$, a bounded set.

Now we need only show the weak-star continuity. By separability of $ell^2$, it is enough to check the convergence of limits. Assume that ${y_j}_{jinmathbb N}xrightarrow{w*} y$. We need to show that $f(y_j)rightarrow f(y)$. Note that by the Banach-Steinhaus theorem, weak-star convergent sequences must be bounded. Pick $R$ such that $|y_j|<R$ for each $j$.

Now let $epsilon>0$. Pick $N$ large so that $sum_{n>N} frac{1}{2^n}R<frac{epsilon}{4}$. Pick $K_1,ldots,K_N$ large (from the weak-star convergence of ${y_j}$) so that
$$forall j>K_i,quad |langle e_i,(y-y_j)rangle|<frac{epsilon}{4N}$$
and let $M=text{max}{N,K_1,ldots,K_N}$. Then for all $j>M$,
begin{align*}
|f(y_j)-f(y)|&=bigg|sum_{ninmathbb N}|langle 2^{-n}e_n,y_jrangle|-sum_{ninmathbb N}|langle 2^{-n}e_n,yrangle|bigg|\
&lebigg|sum_{nle N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|+bigg|sum_{n>N}big[|langle 2^{-n}e_n,y_jrangle|-|langle 2^{-n}e_n,yrangle|big]bigg|\
&le bigg|sum_{nle N}frac{epsilon}{2N}bigg|+bigg|sum_{n>N} 2^{-n}cdot 2Rbigg|\
&< frac{epsilon}{2}+frac{epsilon}{2}\
&<epsilon
end{align*}
so the function is weak-star continuous.

share|cite|improve this answer

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

$endgroup$

share|cite|improve this answer

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516

answered Mar 14 at 16:14

Ashwin TrisalAshwin Trisal

1,2891516