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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          Le Mesnil-Réaume

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          Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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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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          web3.py web3.isConnected() returns false always

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          1 I am trying an tutorial on web3.py so i created a profile in infura and used its project id.But it shows false always when i try web3.isConnected() PS C:UsersHP> python Python 3.6.3 (v3.6.3:2c5fed8, Oct 3 2017, 18:11:49) [MSC v.1900 64 bit (AMD64)] on win32 Type "help", "copyright", "credits" or "license" for more information. >>> from web3 import Web3 >>> infura_url = "mainnet.infura.io/v3/******************************" >>> web3 = Web3(Web3.HTTPProvider(infura_url)) >>> web3.isConnected() False web3.py share | improve this question
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