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











$endgroup$

















    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

      403137




      403137






















          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

          27.9k62447




          27.9k62447












          • $begingroup$
            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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          Probability when a professor distributes a quiz and homework assignment to a class of n students.

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          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
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          Aardman Animations

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          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
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          Are they similar matrix

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          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
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