$operatorname{Ker}(T)$, $T(f) = f'g + fg'$












1












$begingroup$


Let $g in C^{1}[0,1]$



Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$



I have to calculate $operatorname{Ker}(T)$



I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$



If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$



Thus $f = frac{c}{g}$



But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed



Can anybody help me?






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
    $endgroup$
    – Will M.
    Dec 2 '18 at 0:34










  • $begingroup$
    No really, I thought about that but I still do not deduce any expression for f
    $endgroup$
    – ZAF
    Dec 2 '18 at 0:35
















1












$begingroup$


Let $g in C^{1}[0,1]$



Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$



I have to calculate $operatorname{Ker}(T)$



I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$



If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$



Thus $f = frac{c}{g}$



But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed



Can anybody help me?






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
    $endgroup$
    – Will M.
    Dec 2 '18 at 0:34










  • $begingroup$
    No really, I thought about that but I still do not deduce any expression for f
    $endgroup$
    – ZAF
    Dec 2 '18 at 0:35














1












1








1





$begingroup$


Let $g in C^{1}[0,1]$



Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$



I have to calculate $operatorname{Ker}(T)$



I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$



If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$



Thus $f = frac{c}{g}$



But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed



Can anybody help me?






linear-transformations







share|cite|improve this question











$endgroup$




Let $g in C^{1}[0,1]$



Let $T:C^{1}[0,1] to C[0,1]$ linear transformation such that $T(f) = (fg)'$



I have to calculate $operatorname{Ker}(T)$



I know if $f in operatorname{Ker}(T)$ then $(fg)' = 0$ so $fg = c$ in $[0,1]$



If $c not = 0$, then $g(x) not = 0$ $forall x in [0,1]$



Thus $f = frac{c}{g}$



But if there exists $x in [0,1]$ such that $g(x) = 0$ I don't know how to proceed



Can anybody help me?






linear-transformations




linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 0:46









Bernard

119k639112




119k639112










asked Dec 2 '18 at 0:30









ZAFZAF

4357




4357












  • $begingroup$
    If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
    $endgroup$
    – Will M.
    Dec 2 '18 at 0:34










  • $begingroup$
    No really, I thought about that but I still do not deduce any expression for f
    $endgroup$
    – ZAF
    Dec 2 '18 at 0:35


















  • $begingroup$
    If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
    $endgroup$
    – Will M.
    Dec 2 '18 at 0:34










  • $begingroup$
    No really, I thought about that but I still do not deduce any expression for f
    $endgroup$
    – ZAF
    Dec 2 '18 at 0:35
















$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34




$begingroup$
If $g(x)=0$ for some $x,$ then $c = 0.$ IS this of help?
$endgroup$
– Will M.
Dec 2 '18 at 0:34












$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35




$begingroup$
No really, I thought about that but I still do not deduce any expression for f
$endgroup$
– ZAF
Dec 2 '18 at 0:35










2 Answers
2






active

oldest

votes


















0












$begingroup$

If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Noting as you did that
    $$
    fg=c
    $$
    for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
    $$
    ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
    $$
    If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.






          share|cite|improve this answer









          $endgroup$



          If $c=0$ the answer $ker (T)={f:f(x)=0$ on the open set ${y:g(y) neq 0} }$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 0:39









          Kavi Rama MurthyKavi Rama Murthy

          52.8k32055




          52.8k32055























              0












              $begingroup$

              Noting as you did that
              $$
              fg=c
              $$
              for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
              $$
              ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
              $$
              If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Noting as you did that
                $$
                fg=c
                $$
                for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
                $$
                ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
                $$
                If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Noting as you did that
                  $$
                  fg=c
                  $$
                  for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
                  $$
                  ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
                  $$
                  If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.






                  share|cite|improve this answer









                  $endgroup$



                  Noting as you did that
                  $$
                  fg=c
                  $$
                  for some constant $c$, we can deduce that if $g$ vanishes anywhere, $c=0$, and $f$ then is forced to vanish whenever $g$ doesn't, indeed
                  $$
                  ker(T)={ fin C^1[0,1]:;f(x)=0 ;forall x;text{s.t}; g(x)ne 0}
                  $$
                  If $g$ does not vanish, we have the simpler expression $f=frac{c}{g}$ for a constant $c$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 0:41









                  qbertqbert

                  22.1k32460




                  22.1k32460






























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