How the supremum of a function and its derivative in the unit interval are related in normed spaces












2












$begingroup$


Let $E={ f: ,fin C^1([0,1]), f(0)=0}$ and the norm defined on $E$ be $$ left|fright|_E = sup_{xin [0,1]} ,|f'(x)|,,, fin E, $$ and $F=C,([0,1])$ with the norm $$ left|fright|_F = sup_{xin [0,1]} ,|f(x)|,,, fin F.$$



I need to prove that $forall fin E, $ $$ left|fright|_F leq left|fright|_E.$$



Since the interval $[0,1]$ is compact, $fin E, f'in F$ are continuous, the maxima of these functions are attained in the unit intarval. Thus, $exists x_1, x_2in [0,1]$ such that we can write: $$ left|fright|_F= max_{xin [0,1]}|f(x)|=|f(x_1)| ,,text{and},,left|fright|_E= max_{xin [0,1]}|f'(x)|=|f'(x_2)|.$$ Also, $f(x+h)=f(x)+f'(x)h +o(h)$. After taking the norm and using the triangle inequality I can not reach the point which would suggest me the solution.



Can you provide a solution proposal or give me a hint. Thanks.






calculus analysis normed-spaces







share|cite|improve this question











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    2












    $begingroup$


    Let $E={ f: ,fin C^1([0,1]), f(0)=0}$ and the norm defined on $E$ be $$ left|fright|_E = sup_{xin [0,1]} ,|f'(x)|,,, fin E, $$ and $F=C,([0,1])$ with the norm $$ left|fright|_F = sup_{xin [0,1]} ,|f(x)|,,, fin F.$$



    I need to prove that $forall fin E, $ $$ left|fright|_F leq left|fright|_E.$$



    Since the interval $[0,1]$ is compact, $fin E, f'in F$ are continuous, the maxima of these functions are attained in the unit intarval. Thus, $exists x_1, x_2in [0,1]$ such that we can write: $$ left|fright|_F= max_{xin [0,1]}|f(x)|=|f(x_1)| ,,text{and},,left|fright|_E= max_{xin [0,1]}|f'(x)|=|f'(x_2)|.$$ Also, $f(x+h)=f(x)+f'(x)h +o(h)$. After taking the norm and using the triangle inequality I can not reach the point which would suggest me the solution.



    Can you provide a solution proposal or give me a hint. Thanks.






    calculus analysis normed-spaces







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $E={ f: ,fin C^1([0,1]), f(0)=0}$ and the norm defined on $E$ be $$ left|fright|_E = sup_{xin [0,1]} ,|f'(x)|,,, fin E, $$ and $F=C,([0,1])$ with the norm $$ left|fright|_F = sup_{xin [0,1]} ,|f(x)|,,, fin F.$$



      I need to prove that $forall fin E, $ $$ left|fright|_F leq left|fright|_E.$$



      Since the interval $[0,1]$ is compact, $fin E, f'in F$ are continuous, the maxima of these functions are attained in the unit intarval. Thus, $exists x_1, x_2in [0,1]$ such that we can write: $$ left|fright|_F= max_{xin [0,1]}|f(x)|=|f(x_1)| ,,text{and},,left|fright|_E= max_{xin [0,1]}|f'(x)|=|f'(x_2)|.$$ Also, $f(x+h)=f(x)+f'(x)h +o(h)$. After taking the norm and using the triangle inequality I can not reach the point which would suggest me the solution.



      Can you provide a solution proposal or give me a hint. Thanks.






      calculus analysis normed-spaces







      share|cite|improve this question











      $endgroup$




      Let $E={ f: ,fin C^1([0,1]), f(0)=0}$ and the norm defined on $E$ be $$ left|fright|_E = sup_{xin [0,1]} ,|f'(x)|,,, fin E, $$ and $F=C,([0,1])$ with the norm $$ left|fright|_F = sup_{xin [0,1]} ,|f(x)|,,, fin F.$$



      I need to prove that $forall fin E, $ $$ left|fright|_F leq left|fright|_E.$$



      Since the interval $[0,1]$ is compact, $fin E, f'in F$ are continuous, the maxima of these functions are attained in the unit intarval. Thus, $exists x_1, x_2in [0,1]$ such that we can write: $$ left|fright|_F= max_{xin [0,1]}|f(x)|=|f(x_1)| ,,text{and},,left|fright|_E= max_{xin [0,1]}|f'(x)|=|f'(x_2)|.$$ Also, $f(x+h)=f(x)+f'(x)h +o(h)$. After taking the norm and using the triangle inequality I can not reach the point which would suggest me the solution.



      Can you provide a solution proposal or give me a hint. Thanks.






      calculus analysis normed-spaces




      calculus analysis normed-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 20:09







      user249018

















      asked Dec 19 '18 at 19:46









      user249018user249018

      435137




      435137






















          1 Answer
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          $begingroup$

          You were on the right track. Let $x_1 in [0,1]$ such that $|f(x_1)| = |f|_F$.



          If $x_1 = 0$ then $f equiv 0$ so the statement holds.



          If $x_1 > 0$, the mean value theorem implies that $exists c in langle 0, x_1rangle$ such that
          $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$
          so
          $$|f|_F = |f(x_1)| = |f'(c)||x_1| le |f'(c)| le |f|_E$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thanks. It looks so easy and yet I could not make it.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:19










          • $begingroup$
            Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:45






          • 1




            $begingroup$
            @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
            $endgroup$
            – mechanodroid
            Dec 19 '18 at 21:07












          • $begingroup$
            Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
            $endgroup$
            – user249018
            Dec 19 '18 at 22:23










          • $begingroup$
            @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
            $endgroup$
            – mechanodroid
            Dec 20 '18 at 7:25











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          $begingroup$

          You were on the right track. Let $x_1 in [0,1]$ such that $|f(x_1)| = |f|_F$.



          If $x_1 = 0$ then $f equiv 0$ so the statement holds.



          If $x_1 > 0$, the mean value theorem implies that $exists c in langle 0, x_1rangle$ such that
          $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$
          so
          $$|f|_F = |f(x_1)| = |f'(c)||x_1| le |f'(c)| le |f|_E$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thanks. It looks so easy and yet I could not make it.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:19










          • $begingroup$
            Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:45






          • 1




            $begingroup$
            @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
            $endgroup$
            – mechanodroid
            Dec 19 '18 at 21:07












          • $begingroup$
            Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
            $endgroup$
            – user249018
            Dec 19 '18 at 22:23










          • $begingroup$
            @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
            $endgroup$
            – mechanodroid
            Dec 20 '18 at 7:25
















          3












          $begingroup$

          You were on the right track. Let $x_1 in [0,1]$ such that $|f(x_1)| = |f|_F$.



          If $x_1 = 0$ then $f equiv 0$ so the statement holds.



          If $x_1 > 0$, the mean value theorem implies that $exists c in langle 0, x_1rangle$ such that
          $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$
          so
          $$|f|_F = |f(x_1)| = |f'(c)||x_1| le |f'(c)| le |f|_E$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Many thanks. It looks so easy and yet I could not make it.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:19










          • $begingroup$
            Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:45






          • 1




            $begingroup$
            @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
            $endgroup$
            – mechanodroid
            Dec 19 '18 at 21:07












          • $begingroup$
            Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
            $endgroup$
            – user249018
            Dec 19 '18 at 22:23










          • $begingroup$
            @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
            $endgroup$
            – mechanodroid
            Dec 20 '18 at 7:25














          3












          3








          3





          $begingroup$

          You were on the right track. Let $x_1 in [0,1]$ such that $|f(x_1)| = |f|_F$.



          If $x_1 = 0$ then $f equiv 0$ so the statement holds.



          If $x_1 > 0$, the mean value theorem implies that $exists c in langle 0, x_1rangle$ such that
          $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$
          so
          $$|f|_F = |f(x_1)| = |f'(c)||x_1| le |f'(c)| le |f|_E$$






          share|cite|improve this answer









          $endgroup$



          You were on the right track. Let $x_1 in [0,1]$ such that $|f(x_1)| = |f|_F$.



          If $x_1 = 0$ then $f equiv 0$ so the statement holds.



          If $x_1 > 0$, the mean value theorem implies that $exists c in langle 0, x_1rangle$ such that
          $$f(x_1) - f(0) = f'(c)(x_1 - 0)$$
          so
          $$|f|_F = |f(x_1)| = |f'(c)||x_1| le |f'(c)| le |f|_E$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 20:00









          mechanodroidmechanodroid

          28.9k62548




          28.9k62548












          • $begingroup$
            Many thanks. It looks so easy and yet I could not make it.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:19










          • $begingroup$
            Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:45






          • 1




            $begingroup$
            @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
            $endgroup$
            – mechanodroid
            Dec 19 '18 at 21:07












          • $begingroup$
            Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
            $endgroup$
            – user249018
            Dec 19 '18 at 22:23










          • $begingroup$
            @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
            $endgroup$
            – mechanodroid
            Dec 20 '18 at 7:25


















          • $begingroup$
            Many thanks. It looks so easy and yet I could not make it.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:19










          • $begingroup$
            Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
            $endgroup$
            – user249018
            Dec 19 '18 at 20:45






          • 1




            $begingroup$
            @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
            $endgroup$
            – mechanodroid
            Dec 19 '18 at 21:07












          • $begingroup$
            Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
            $endgroup$
            – user249018
            Dec 19 '18 at 22:23










          • $begingroup$
            @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
            $endgroup$
            – mechanodroid
            Dec 20 '18 at 7:25
















          $begingroup$
          Many thanks. It looks so easy and yet I could not make it.
          $endgroup$
          – user249018
          Dec 19 '18 at 20:19




          $begingroup$
          Many thanks. It looks so easy and yet I could not make it.
          $endgroup$
          – user249018
          Dec 19 '18 at 20:19












          $begingroup$
          Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
          $endgroup$
          – user249018
          Dec 19 '18 at 20:45




          $begingroup$
          Under the same conditions can you give me a hint to prove that the function $s(f)=f'+f²$ is differentiable.
          $endgroup$
          – user249018
          Dec 19 '18 at 20:45




          1




          1




          $begingroup$
          @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
          $endgroup$
          – mechanodroid
          Dec 19 '18 at 21:07






          $begingroup$
          @user249018 The differential of $s : E to F$ at point $f in E$ should be given by $Ds(f)h := h' + 2fh$. Indeed, it is bounded with $|Ds(f)| le 1+2|f|_E$ and begin{align} lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}&= lim_{hto 0}frac{|s(f+h) - s(f) - Ds(f)h|_F}{|h|_E}\ &= lim_{hto 0}frac{|f'+h' +(f+h)^2- f'-f^2 - h'-2fh|_F}{|h|_E}\ &= lim_{hto 0}frac{|h^2|_F}{|h|_E}\ &= lim_{hto 0}frac{|h|_F^2}{|h|_E}\ &le lim_{hto 0}frac{|h|_E^2}{|h|_E}\ &= lim_{hto 0} |h|_E\ &= 0 end{align}
          $endgroup$
          – mechanodroid
          Dec 19 '18 at 21:07














          $begingroup$
          Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
          $endgroup$
          – user249018
          Dec 19 '18 at 22:23




          $begingroup$
          Thanks. How you come to define the differential in such a way ? I tried it with $Ds(f):=2f$ but did not work, thats why I needed some support. Is it very obvious or one should try different options such that the rest of the first order Taylor development become negligeable in respect to the function $h$ ? Anyway thank you for the very clear answer.
          $endgroup$
          – user249018
          Dec 19 '18 at 22:23












          $begingroup$
          @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
          $endgroup$
          – mechanodroid
          Dec 20 '18 at 7:25




          $begingroup$
          @user249018 I tried to define it such that $s(f+h) - s(f) - Ds(f)h = o(h^2)$, it was mostly trial and error.
          $endgroup$
          – mechanodroid
          Dec 20 '18 at 7:25


















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          Bundesstraße 106

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          Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
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          Verónica Boquete

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          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
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          Ida-Boy-Ed-Garten

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          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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