Is the set ${|f(0)|: int_{0}^{1}|f(t)|dtle1}$ bounded?












4














Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:



Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?



I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.



But how can I prove this formally?






calculus real-analysis functional-analysis analysis







share|cite|improve this question
























  • As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
    – Alex R.
    Nov 29 at 21:32












  • This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
    – Yanko
    Nov 29 at 21:32










  • Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
    – rubikscube09
    Nov 29 at 21:35












  • Sorry for the confusion. Just edited the description to make it clearer.
    – bbw
    Nov 29 at 21:37






  • 1




    Is it really a problem? What about a constant coefficient?
    – Jean-Claude Arbaut
    Nov 29 at 21:51
















4














Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:



Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?



I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.



But how can I prove this formally?






calculus real-analysis functional-analysis analysis







share|cite|improve this question
























  • As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
    – Alex R.
    Nov 29 at 21:32












  • This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
    – Yanko
    Nov 29 at 21:32










  • Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
    – rubikscube09
    Nov 29 at 21:35












  • Sorry for the confusion. Just edited the description to make it clearer.
    – bbw
    Nov 29 at 21:37






  • 1




    Is it really a problem? What about a constant coefficient?
    – Jean-Claude Arbaut
    Nov 29 at 21:51














4












4








4







Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:



Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?



I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.



But how can I prove this formally?






calculus real-analysis functional-analysis analysis







share|cite|improve this question















Let $x_0 in [0,1]$ and define $T:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$. Let $||cdot||_1$ be a norm on $C[0,1]$. Is $T_0$ bounded or not? That is, is the set
$$
left{|T_{0}(f)|:||f||_1 leq 1right}={|f(0)|:||f||_1 leq 1,f in C[0,1]}
$$ bounded? Since $||f||_1:=int_{0}^{1}|f(t)|dt$, the question may be equivalent to the following:



Let $f:[0,1] rightarrow mathbb{R}$ be continuous. Is the set $$left{|f(0)|: int_{0}^{1}|f(t)|dt leq 1right}$$ bounded?



I guess the answer is no. Because, for example, we can have a function whose graph is a narrow spike at the origin but with infinite height. The area enclosed by the graph may be 1 but the value at the origin $f(0)$ which is its height is infinite.



But how can I prove this formally?






calculus real-analysis functional-analysis analysis




calculus real-analysis functional-analysis analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 21:45









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Nov 29 at 21:30









bbw

47038




47038












  • As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
    – Alex R.
    Nov 29 at 21:32












  • This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
    – Yanko
    Nov 29 at 21:32










  • Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
    – rubikscube09
    Nov 29 at 21:35












  • Sorry for the confusion. Just edited the description to make it clearer.
    – bbw
    Nov 29 at 21:37






  • 1




    Is it really a problem? What about a constant coefficient?
    – Jean-Claude Arbaut
    Nov 29 at 21:51


















  • As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
    – Alex R.
    Nov 29 at 21:32












  • This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
    – Yanko
    Nov 29 at 21:32










  • Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
    – rubikscube09
    Nov 29 at 21:35












  • Sorry for the confusion. Just edited the description to make it clearer.
    – bbw
    Nov 29 at 21:37






  • 1




    Is it really a problem? What about a constant coefficient?
    – Jean-Claude Arbaut
    Nov 29 at 21:51
















As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32






As written, this makes little sense. What is the precise definition of the set? There's no condition placed on anything. Is it a set of functions?
– Alex R.
Nov 29 at 21:32














This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32




This is not a set... you need to assume something about $int_0^1 | f(t)|dt$. I mean it should be a true/false statement...
– Yanko
Nov 29 at 21:32












Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35






Is the RHS trying to say that $f$ is absolutely integrable on $[0,1]$? If so, you have the right idea, consider the function $1/sqrt{x}$ and its integral on $[0,1]$
– rubikscube09
Nov 29 at 21:35














Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37




Sorry for the confusion. Just edited the description to make it clearer.
– bbw
Nov 29 at 21:37




1




1




Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51




Is it really a problem? What about a constant coefficient?
– Jean-Claude Arbaut
Nov 29 at 21:51










4 Answers
4






active

oldest

votes


















2














No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.






share|cite|improve this answer





























    1














    We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.



    For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.






    share|cite|improve this answer





























      0














      Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$






      share|cite|improve this answer





























        0














        For every $a>0$, the function
        $$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.



        For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.






        share|cite|improve this answer





















          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019246%2fis-the-set-f0-int-01ftdt-le1-bounded%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.






          share|cite|improve this answer


























            2














            No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.






            share|cite|improve this answer
























              2












              2








              2






              No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.






              share|cite|improve this answer












              No, it is not bounded. Define, for each $ninmathbb N$,$$begin{array}{rccc}f_ncolon&[0,1]&longrightarrow&mathbb R\&t&mapsto&begin{cases}n-n^2t&text{ if }tleqslantfrac1n\0&text{ otherwise.}end{cases}end{array}$$Then$$int_0^1bigllvert f_n(t)bigrrvert,mathrm dt=frac12,$$but $f_n(0)=n$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 29 at 21:50









              José Carlos Santos

              149k22117219




              149k22117219























                  1














                  We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.



                  For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.






                  share|cite|improve this answer


























                    1














                    We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.



                    For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.






                    share|cite|improve this answer
























                      1












                      1








                      1






                      We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.



                      For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.






                      share|cite|improve this answer












                      We can take a function $f_n$ such that $f_n$ is affine on $(0,1/n)$, $f_n(0)=2n$ and $f_n(1/n)=0$ and $0$ for the other values of the interval. Then $leftlVert f_nrightrVert_1=1$.



                      For a formal example: define for each positive integer $n$ the function $f_n$ in the following way: $f_n(t)=-2n^2t+2n$ for $0leqslant tleqslant 1/n$ and $f_n(t)=0$ for $1/nlt tleqslant 1$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 29 at 21:51









                      Davide Giraudo

                      125k16150259




                      125k16150259























                          0














                          Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$






                          share|cite|improve this answer


























                            0














                            Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$






                            share|cite|improve this answer
























                              0












                              0








                              0






                              Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$






                              share|cite|improve this answer












                              Consider the function $f_n(x)=2n(1/n-x)1_{[0,1/n]}(x).$ Then $int_0^1 |f_n(x)|dx=1$ but $f_n(0)=2n.$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Nov 29 at 21:52









                              John_Wick

                              1,356111




                              1,356111























                                  0














                                  For every $a>0$, the function
                                  $$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.



                                  For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.






                                  share|cite|improve this answer


























                                    0














                                    For every $a>0$, the function
                                    $$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.



                                    For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      For every $a>0$, the function
                                      $$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.



                                      For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.






                                      share|cite|improve this answer












                                      For every $a>0$, the function
                                      $$f_a(x)=frac{2 a e^{-a^2 x^2}}{sqrt{pi} textrm{erf}(a)}$$ with the error function $textrm{erf}(a)=frac{2}{pi}int_0^a e^{-t^2}dt$ is in the set ${|f(0)|: int_{0}^{1}|f(t)|dt=1}$ and evaluates to $f_a(0)=frac{2 a }{sqrt{pi} textrm{erf}(a)}$. Because $textrm{erf}(a) rightarrow 1$ as $arightarrow infty$, we obtain $f_a(0)$ arbitrarily large as we increase $a$. As a consequence, $f_a(0)$ and thus your set are unbounded.



                                      For every $a>0$, the function $f_a(x)$ is in $C[0,1]$ and even infinitely differentiable on $[0,1]$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 29 at 21:53









                                      M1183

                                      943




                                      943






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.





                                          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                          Please pay close attention to the following guidance:


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019246%2fis-the-set-f0-int-01ftdt-le1-bounded%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Wiesbaden

                                          Image
                                          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
                                          Read more

                                          Marschland

                                          Image
                                          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
                                          Read more

                                          Dieringhausen

                                          Image
                                          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
                                          Read more