Proving that $exists f in X^*$ : $f(x) = |x|^2$ and $|f| = |x|$












3












$begingroup$


Exercise :




Let $X$ be a normed space. Prove that for all $x in X$ there exists $f in X^*$, such that $f(x) = |x|^2$ and $ |f| = |x |$.




Thoughts :



I apologise for not providing a proper attempt but this is one of the first such exercises I'm handling, so I seem at loss.



Initially, I thought about the Riesz Representation Theorem, which could yield the result straightforward, but the space we are working over must be a Hilbert Space, which we do not know in the given exercise.



The second possible solution could (and probably should) be based on the Hahn-Banach Theorem (or one of its results/applications) but I cannot see a way out.



Any hints or elaborations will be greatly appreciated.






functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Exercise :




    Let $X$ be a normed space. Prove that for all $x in X$ there exists $f in X^*$, such that $f(x) = |x|^2$ and $ |f| = |x |$.




    Thoughts :



    I apologise for not providing a proper attempt but this is one of the first such exercises I'm handling, so I seem at loss.



    Initially, I thought about the Riesz Representation Theorem, which could yield the result straightforward, but the space we are working over must be a Hilbert Space, which we do not know in the given exercise.



    The second possible solution could (and probably should) be based on the Hahn-Banach Theorem (or one of its results/applications) but I cannot see a way out.



    Any hints or elaborations will be greatly appreciated.






    functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      Exercise :




      Let $X$ be a normed space. Prove that for all $x in X$ there exists $f in X^*$, such that $f(x) = |x|^2$ and $ |f| = |x |$.




      Thoughts :



      I apologise for not providing a proper attempt but this is one of the first such exercises I'm handling, so I seem at loss.



      Initially, I thought about the Riesz Representation Theorem, which could yield the result straightforward, but the space we are working over must be a Hilbert Space, which we do not know in the given exercise.



      The second possible solution could (and probably should) be based on the Hahn-Banach Theorem (or one of its results/applications) but I cannot see a way out.



      Any hints or elaborations will be greatly appreciated.






      functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem







      share|cite|improve this question











      $endgroup$




      Exercise :




      Let $X$ be a normed space. Prove that for all $x in X$ there exists $f in X^*$, such that $f(x) = |x|^2$ and $ |f| = |x |$.




      Thoughts :



      I apologise for not providing a proper attempt but this is one of the first such exercises I'm handling, so I seem at loss.



      Initially, I thought about the Riesz Representation Theorem, which could yield the result straightforward, but the space we are working over must be a Hilbert Space, which we do not know in the given exercise.



      The second possible solution could (and probably should) be based on the Hahn-Banach Theorem (or one of its results/applications) but I cannot see a way out.



      Any hints or elaborations will be greatly appreciated.






      functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem




      functional-analysis normed-spaces riesz-representation-theorem hahn-banach-theorem






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 15:20







      Rebellos

















      asked Dec 17 '18 at 14:18









      RebellosRebellos

      14.8k31248




      14.8k31248






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          If $x=0$, we are done. Now let $x in X$ and $x ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g in X^*$ with



          $g(x)=||x||$ and $||g||=1.$



          Now it is easy to see that $f:=||x||g$ has the desired properties.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Hint:



            Fix an $xin X$. If $x=0$ then the answer is rather easy, so we can assume $xneq 0$.



            Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:={ alpha cdot x | alpha in mathbb R}$.
            Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
              $endgroup$
              – Rebellos
              Dec 17 '18 at 14:33












            • $begingroup$
              I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
              $endgroup$
              – Rebellos
              Dec 17 '18 at 14:43










            • $begingroup$
              No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
              $endgroup$
              – supinf
              Dec 17 '18 at 14:46










            • $begingroup$
              $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
              $endgroup$
              – Rebellos
              Dec 17 '18 at 14:46












            • $begingroup$
              You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
              $endgroup$
              – supinf
              Dec 17 '18 at 14:54













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            2 Answers
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            active

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            2 Answers
            2






            active

            oldest

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            active

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            votes






            active

            oldest

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            3












            $begingroup$

            If $x=0$, we are done. Now let $x in X$ and $x ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g in X^*$ with



            $g(x)=||x||$ and $||g||=1.$



            Now it is easy to see that $f:=||x||g$ has the desired properties.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              If $x=0$, we are done. Now let $x in X$ and $x ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g in X^*$ with



              $g(x)=||x||$ and $||g||=1.$



              Now it is easy to see that $f:=||x||g$ has the desired properties.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                If $x=0$, we are done. Now let $x in X$ and $x ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g in X^*$ with



                $g(x)=||x||$ and $||g||=1.$



                Now it is easy to see that $f:=||x||g$ has the desired properties.






                share|cite|improve this answer









                $endgroup$



                If $x=0$, we are done. Now let $x in X$ and $x ne 0$. A consequence of the Hahn-Banach theorem is the existence of some $g in X^*$ with



                $g(x)=||x||$ and $||g||=1.$



                Now it is easy to see that $f:=||x||g$ has the desired properties.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 14:29









                FredFred

                46.9k1848




                46.9k1848























                    2












                    $begingroup$

                    Hint:



                    Fix an $xin X$. If $x=0$ then the answer is rather easy, so we can assume $xneq 0$.



                    Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:={ alpha cdot x | alpha in mathbb R}$.
                    Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:33












                    • $begingroup$
                      I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:43










                    • $begingroup$
                      No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:46










                    • $begingroup$
                      $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:46












                    • $begingroup$
                      You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:54


















                    2












                    $begingroup$

                    Hint:



                    Fix an $xin X$. If $x=0$ then the answer is rather easy, so we can assume $xneq 0$.



                    Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:={ alpha cdot x | alpha in mathbb R}$.
                    Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:33












                    • $begingroup$
                      I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:43










                    • $begingroup$
                      No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:46










                    • $begingroup$
                      $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:46












                    • $begingroup$
                      You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:54
















                    2












                    2








                    2





                    $begingroup$

                    Hint:



                    Fix an $xin X$. If $x=0$ then the answer is rather easy, so we can assume $xneq 0$.



                    Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:={ alpha cdot x | alpha in mathbb R}$.
                    Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.






                    share|cite|improve this answer











                    $endgroup$



                    Hint:



                    Fix an $xin X$. If $x=0$ then the answer is rather easy, so we can assume $xneq 0$.



                    Next, define a suitable functional on the linear hull of $x$, which is the linear subspace $V:={ alpha cdot x | alpha in mathbb R}$.
                    Then extend this functional onto the whole space $X$ using the Hahn-Banach theorem.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 17 '18 at 14:41

























                    answered Dec 17 '18 at 14:21









                    supinfsupinf

                    6,3561028




                    6,3561028












                    • $begingroup$
                      Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:33












                    • $begingroup$
                      I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:43










                    • $begingroup$
                      No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:46










                    • $begingroup$
                      $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:46












                    • $begingroup$
                      You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:54




















                    • $begingroup$
                      Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:33












                    • $begingroup$
                      I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:43










                    • $begingroup$
                      No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:46










                    • $begingroup$
                      $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                      $endgroup$
                      – Rebellos
                      Dec 17 '18 at 14:46












                    • $begingroup$
                      You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                      $endgroup$
                      – supinf
                      Dec 17 '18 at 14:54


















                    $begingroup$
                    Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:33






                    $begingroup$
                    Let's say $f : langle x : x in X rangle to mathbb R$ with $f(y) = langle y,x rangle$. Then we can see that $f(x) = |x|^2$ and $frac{|f(x)|}{|x|} = |x| implies |f| = |x|$. But $ langle x : x in X rangle subseteq X $ and one can see that the $f$ defined is a bounded linear functional, thus it can be extended to $X$ which implies that $exists f in X^*$ such that $f(x) = |x|^2$ and $|f| = |x|$. Does that make sense or do I have any errors ?
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:33














                    $begingroup$
                    I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:43




                    $begingroup$
                    I just saw your edit. I would appreciate an input regarding my attempted approach and/or an input regarding an approach you would recommend.
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:43












                    $begingroup$
                    No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                    $endgroup$
                    – supinf
                    Dec 17 '18 at 14:46




                    $begingroup$
                    No, your approach does not work since $langle cdot,cdotrangle$ is not defined on $X$. $X$ is only a normed vector space in your question. Also the notation $<x : xin X>$ is not a standard notation. if it is a set it would use ${}$ instead of $<>$. Note that $X={x:xin X}$, so your space is not a proper subspace of $X$.
                    $endgroup$
                    – supinf
                    Dec 17 '18 at 14:46












                    $begingroup$
                    $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:46






                    $begingroup$
                    $langle rangle$ denotes the linear hull. If my approach is not correct, would you mind elaborating as such, in order to properly understand ? I guess another proper functional will be $f : {lambda cdot x | x in X, ; lambda in mathbb R} to mathbb R$ such that $f(x) = frac{1}{lambda^2}|x|^2$ ?
                    $endgroup$
                    – Rebellos
                    Dec 17 '18 at 14:46














                    $begingroup$
                    You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                    $endgroup$
                    – supinf
                    Dec 17 '18 at 14:54






                    $begingroup$
                    You should work with a fixed $x$ and not use it as a variable for the definition of functions or sets. Note that your suggestion for $f$ is not linear. With the linear hull notation, the space should be $V=langle x rangle$, which has dimension $1$.
                    $endgroup$
                    – supinf
                    Dec 17 '18 at 14:54




















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                    Wiesbaden

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                    Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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                    Marschland

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                    Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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                    Dieringhausen

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                    Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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