Why an unbounded operator is not-constructive relying on Hahn Banach theorem?












4












$begingroup$


Let $T: mathcal S(mathbb R)to L^2(mathbb R)$ defined by $$Tf(x)=f'(x),$$
where $mathcal S(mathbb R)$ is the Schwarz space on $mathbb R$. The question is : is there a continuous extension to $L^2(mathbb R)$, and the answer is no, because $frac{|Tf_k|}{|f_k|}to infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.



You'll see the example here



1) Why is this a counter-example since $e^{-k|x|}notin mathcal S(mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $mathbb R$, is it ? Because on $L^2(mathbb R)setminus mathcal S(mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $mathcal S(mathbb R)$ than on $L^2(mathbb R)$.



2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).






functional-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
    $endgroup$
    – Surb
    Jan 6 at 16:07
















4












$begingroup$


Let $T: mathcal S(mathbb R)to L^2(mathbb R)$ defined by $$Tf(x)=f'(x),$$
where $mathcal S(mathbb R)$ is the Schwarz space on $mathbb R$. The question is : is there a continuous extension to $L^2(mathbb R)$, and the answer is no, because $frac{|Tf_k|}{|f_k|}to infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.



You'll see the example here



1) Why is this a counter-example since $e^{-k|x|}notin mathcal S(mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $mathbb R$, is it ? Because on $L^2(mathbb R)setminus mathcal S(mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $mathcal S(mathbb R)$ than on $L^2(mathbb R)$.



2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).






functional-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
    $endgroup$
    – Surb
    Jan 6 at 16:07














4












4








4


2



$begingroup$


Let $T: mathcal S(mathbb R)to L^2(mathbb R)$ defined by $$Tf(x)=f'(x),$$
where $mathcal S(mathbb R)$ is the Schwarz space on $mathbb R$. The question is : is there a continuous extension to $L^2(mathbb R)$, and the answer is no, because $frac{|Tf_k|}{|f_k|}to infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.



You'll see the example here



1) Why is this a counter-example since $e^{-k|x|}notin mathcal S(mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $mathbb R$, is it ? Because on $L^2(mathbb R)setminus mathcal S(mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $mathcal S(mathbb R)$ than on $L^2(mathbb R)$.



2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).






functional-analysis







share|cite|improve this question









$endgroup$




Let $T: mathcal S(mathbb R)to L^2(mathbb R)$ defined by $$Tf(x)=f'(x),$$
where $mathcal S(mathbb R)$ is the Schwarz space on $mathbb R$. The question is : is there a continuous extension to $L^2(mathbb R)$, and the answer is no, because $frac{|Tf_k|}{|f_k|}to infty $ where $f_k(x)=e^{-k|x|},$ and thus $T$ is unbounded.



You'll see the example here



1) Why is this a counter-example since $e^{-k|x|}notin mathcal S(mathbb R)$ ? and is even not derivable at $0$, so $f'(x)$ is not well defined on $mathbb R$, is it ? Because on $L^2(mathbb R)setminus mathcal S(mathbb R)$, the operator $T$ can be different than $Tf(x)=f'(x)$ ? (for example the fourier transform is not the same on $mathcal S(mathbb R)$ than on $L^2(mathbb R)$.



2) Just under the example in the previous link, they say that the existence of unbounded operators defined everywhere is non constructive relying on Hahn-Banach theorem. Could someone explain what this is true ? I guess that here unbounded mean not continuous ? (and not defined on a subspace of the domain).






functional-analysis




functional-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 15:47









NewMathNewMath

4059




4059












  • $begingroup$
    If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
    $endgroup$
    – Surb
    Jan 6 at 16:07


















  • $begingroup$
    If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
    $endgroup$
    – Surb
    Jan 6 at 16:07
















$begingroup$
If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
$endgroup$
– Surb
Jan 6 at 16:07




$begingroup$
If $f$ is derivable a.e. and $f'in L^2$, then $Tf=f'$ is well defined. But as your example show, $T$ won't be bounded on this more general set.
$endgroup$
– Surb
Jan 6 at 16:07










1 Answer
1






active

oldest

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6












$begingroup$

By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
    $endgroup$
    – NewMath
    Jan 6 at 16:12








  • 1




    $begingroup$
    It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
    $endgroup$
    – Robert Israel
    Jan 6 at 19:50











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active

oldest

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6












$begingroup$

By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
    $endgroup$
    – NewMath
    Jan 6 at 16:12








  • 1




    $begingroup$
    It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
    $endgroup$
    – Robert Israel
    Jan 6 at 19:50
















6












$begingroup$

By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
    $endgroup$
    – NewMath
    Jan 6 at 16:12








  • 1




    $begingroup$
    It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
    $endgroup$
    – Robert Israel
    Jan 6 at 19:50














6












6








6





$begingroup$

By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.






share|cite|improve this answer











$endgroup$



By a theorem of Solovay, it's consistent with ZF (the Zermelo-Frankel axioms, without Choice) + DC (Dependent Choice) that every subset of a complete separable metric space has the property of Baire. From this it would follow that every everywhere-defined linear operator from a Hilbert space to itself is continuous.
So to get an unbounded operator you need Axiom of Choice (or something pretty close to it). See J.D.M. Wright, BAMS 79 (1973) 1247-1250.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 19:51

























answered Jan 6 at 16:06









Robert IsraelRobert Israel

329k23217470




329k23217470








  • 1




    $begingroup$
    So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
    $endgroup$
    – NewMath
    Jan 6 at 16:12








  • 1




    $begingroup$
    It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
    $endgroup$
    – Robert Israel
    Jan 6 at 19:50














  • 1




    $begingroup$
    So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
    $endgroup$
    – NewMath
    Jan 6 at 16:12








  • 1




    $begingroup$
    It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
    $endgroup$
    – Robert Israel
    Jan 6 at 19:50








1




1




$begingroup$
So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
$endgroup$
– NewMath
Jan 6 at 16:12






$begingroup$
So if $T: L^2to L^2$, then $T$ is always continuous ? But this mean that there are non unbounded operator on a hilbert space that is defined every where...
$endgroup$
– NewMath
Jan 6 at 16:12






1




1




$begingroup$
It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
$endgroup$
– Robert Israel
Jan 6 at 19:50




$begingroup$
It is consistent with ZF+DC that every $T: L^2 to L^2$ is continuous. Axiom of Choice says there are unbounded operators, but without Axiom of Choice you can't construct one and prove it is unbounded.
$endgroup$
– Robert Israel
Jan 6 at 19:50


















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