Proving $f$ zero function if $|f'(x)|le A|f(x)|$












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This is baby Rudin exer 5.26, and the solution I have:




Exercise: Suppose $f$ is differentiable on $[a,b]$, $f(a)=0$, and there is a real number $A$ such that $|f'(x)|le A|f(x)|$ on $[a,b]$. Prove that $f(x)=0$ for all $[a,b].$



Solution: Suppose there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d.$ By passing to consideration of $-f(x)$ if necessary, we can assume $f(x)>0$ for $c<x<d$. The function $g(x)=ln f(x)$ is then defined for $c<x<d$ and its derivative satisfies $$|g'(x)|=|frac{f'(x)}{f(x)}|le A.$$ The mean value theorem then implies $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$ for all $xin(c,d)$. But this is contradiction since, $g(x)to-infty$ as $xto c$.




I can't understand how did we get $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$.



I see that by MVT, $$g(x)-gleft(frac{c+d}{2}right)=g'(y)left(x-frac{c+d}{2}right)le g'(y)left(d-frac{c+d}{2}right)=g'(y)left(frac{d-c}{2}right)$$ for some $yin (c,d)$. Hence, $$left|left|g(x)right|-left|gleft(frac{c+d}{2}right)right|right|le Aleft(frac{d-c}{2}right),$$ but, that's not desired inequality.






real-analysis derivatives inequality proof-explanation







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$endgroup$

















    0












    $begingroup$


    This is baby Rudin exer 5.26, and the solution I have:




    Exercise: Suppose $f$ is differentiable on $[a,b]$, $f(a)=0$, and there is a real number $A$ such that $|f'(x)|le A|f(x)|$ on $[a,b]$. Prove that $f(x)=0$ for all $[a,b].$



    Solution: Suppose there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d.$ By passing to consideration of $-f(x)$ if necessary, we can assume $f(x)>0$ for $c<x<d$. The function $g(x)=ln f(x)$ is then defined for $c<x<d$ and its derivative satisfies $$|g'(x)|=|frac{f'(x)}{f(x)}|le A.$$ The mean value theorem then implies $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$ for all $xin(c,d)$. But this is contradiction since, $g(x)to-infty$ as $xto c$.




    I can't understand how did we get $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$.



    I see that by MVT, $$g(x)-gleft(frac{c+d}{2}right)=g'(y)left(x-frac{c+d}{2}right)le g'(y)left(d-frac{c+d}{2}right)=g'(y)left(frac{d-c}{2}right)$$ for some $yin (c,d)$. Hence, $$left|left|g(x)right|-left|gleft(frac{c+d}{2}right)right|right|le Aleft(frac{d-c}{2}right),$$ but, that's not desired inequality.






    real-analysis derivatives inequality proof-explanation







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      0



      $begingroup$


      This is baby Rudin exer 5.26, and the solution I have:




      Exercise: Suppose $f$ is differentiable on $[a,b]$, $f(a)=0$, and there is a real number $A$ such that $|f'(x)|le A|f(x)|$ on $[a,b]$. Prove that $f(x)=0$ for all $[a,b].$



      Solution: Suppose there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d.$ By passing to consideration of $-f(x)$ if necessary, we can assume $f(x)>0$ for $c<x<d$. The function $g(x)=ln f(x)$ is then defined for $c<x<d$ and its derivative satisfies $$|g'(x)|=|frac{f'(x)}{f(x)}|le A.$$ The mean value theorem then implies $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$ for all $xin(c,d)$. But this is contradiction since, $g(x)to-infty$ as $xto c$.




      I can't understand how did we get $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$.



      I see that by MVT, $$g(x)-gleft(frac{c+d}{2}right)=g'(y)left(x-frac{c+d}{2}right)le g'(y)left(d-frac{c+d}{2}right)=g'(y)left(frac{d-c}{2}right)$$ for some $yin (c,d)$. Hence, $$left|left|g(x)right|-left|gleft(frac{c+d}{2}right)right|right|le Aleft(frac{d-c}{2}right),$$ but, that's not desired inequality.






      real-analysis derivatives inequality proof-explanation







      share|cite|improve this question











      $endgroup$




      This is baby Rudin exer 5.26, and the solution I have:




      Exercise: Suppose $f$ is differentiable on $[a,b]$, $f(a)=0$, and there is a real number $A$ such that $|f'(x)|le A|f(x)|$ on $[a,b]$. Prove that $f(x)=0$ for all $[a,b].$



      Solution: Suppose there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d.$ By passing to consideration of $-f(x)$ if necessary, we can assume $f(x)>0$ for $c<x<d$. The function $g(x)=ln f(x)$ is then defined for $c<x<d$ and its derivative satisfies $$|g'(x)|=|frac{f'(x)}{f(x)}|le A.$$ The mean value theorem then implies $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$ for all $xin(c,d)$. But this is contradiction since, $g(x)to-infty$ as $xto c$.




      I can't understand how did we get $g(x)ge gleft(frac{c+d}{2}right)-Aleft(frac{d-c}{2}right)$.



      I see that by MVT, $$g(x)-gleft(frac{c+d}{2}right)=g'(y)left(x-frac{c+d}{2}right)le g'(y)left(d-frac{c+d}{2}right)=g'(y)left(frac{d-c}{2}right)$$ for some $yin (c,d)$. Hence, $$left|left|g(x)right|-left|gleft(frac{c+d}{2}right)right|right|le Aleft(frac{d-c}{2}right),$$ but, that's not desired inequality.






      real-analysis derivatives inequality proof-explanation




      real-analysis derivatives inequality proof-explanation






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      edited Dec 2 '18 at 13:24









      José Carlos Santos

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      asked Dec 2 '18 at 12:57









      SilentSilent

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          $begingroup$

          If $xin(c,d)setminusleft{frac{c+d}2right}$, then$$frac{g(x)-gleft(frac{c+d}2right)}{x-frac{c+d}2}in[-A,A].$$Now, if $xinleft(frac{c+d}2,dright)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant(-A)timesleft(x-frac{c+d}2right)\&geqslant(-A)timesleft(d-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}And if $xinleft(c,frac{c+d}2right)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant Atimesleft(x-frac{c+d}2right)\&geqslant Atimesleft(c-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}






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          • $begingroup$
            thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
            $endgroup$
            – Silent
            Dec 2 '18 at 13:26










          • $begingroup$
            Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
            $endgroup$
            – Silent
            Dec 2 '18 at 13:40










          • $begingroup$
            Yes, that is correct.
            $endgroup$
            – José Carlos Santos
            Dec 2 '18 at 14:25










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Silent
            Dec 2 '18 at 14:27











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          $begingroup$

          If $xin(c,d)setminusleft{frac{c+d}2right}$, then$$frac{g(x)-gleft(frac{c+d}2right)}{x-frac{c+d}2}in[-A,A].$$Now, if $xinleft(frac{c+d}2,dright)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant(-A)timesleft(x-frac{c+d}2right)\&geqslant(-A)timesleft(d-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}And if $xinleft(c,frac{c+d}2right)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant Atimesleft(x-frac{c+d}2right)\&geqslant Atimesleft(c-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
            $endgroup$
            – Silent
            Dec 2 '18 at 13:26










          • $begingroup$
            Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
            $endgroup$
            – Silent
            Dec 2 '18 at 13:40










          • $begingroup$
            Yes, that is correct.
            $endgroup$
            – José Carlos Santos
            Dec 2 '18 at 14:25










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Silent
            Dec 2 '18 at 14:27
















          2












          $begingroup$

          If $xin(c,d)setminusleft{frac{c+d}2right}$, then$$frac{g(x)-gleft(frac{c+d}2right)}{x-frac{c+d}2}in[-A,A].$$Now, if $xinleft(frac{c+d}2,dright)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant(-A)timesleft(x-frac{c+d}2right)\&geqslant(-A)timesleft(d-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}And if $xinleft(c,frac{c+d}2right)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant Atimesleft(x-frac{c+d}2right)\&geqslant Atimesleft(c-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
            $endgroup$
            – Silent
            Dec 2 '18 at 13:26










          • $begingroup$
            Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
            $endgroup$
            – Silent
            Dec 2 '18 at 13:40










          • $begingroup$
            Yes, that is correct.
            $endgroup$
            – José Carlos Santos
            Dec 2 '18 at 14:25










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Silent
            Dec 2 '18 at 14:27














          2












          2








          2





          $begingroup$

          If $xin(c,d)setminusleft{frac{c+d}2right}$, then$$frac{g(x)-gleft(frac{c+d}2right)}{x-frac{c+d}2}in[-A,A].$$Now, if $xinleft(frac{c+d}2,dright)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant(-A)timesleft(x-frac{c+d}2right)\&geqslant(-A)timesleft(d-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}And if $xinleft(c,frac{c+d}2right)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant Atimesleft(x-frac{c+d}2right)\&geqslant Atimesleft(c-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}






          share|cite|improve this answer









          $endgroup$



          If $xin(c,d)setminusleft{frac{c+d}2right}$, then$$frac{g(x)-gleft(frac{c+d}2right)}{x-frac{c+d}2}in[-A,A].$$Now, if $xinleft(frac{c+d}2,dright)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant(-A)timesleft(x-frac{c+d}2right)\&geqslant(-A)timesleft(d-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}And if $xinleft(c,frac{c+d}2right)$, this implies thatbegin{align}g(x)-gleft(frac{c+d}2right)&geqslant Atimesleft(x-frac{c+d}2right)\&geqslant Atimesleft(c-frac{c+d}2right)\&=-Atimesfrac{d-c}2.end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 13:15









          José Carlos SantosJosé Carlos Santos

          155k22124227




          155k22124227












          • $begingroup$
            thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
            $endgroup$
            – Silent
            Dec 2 '18 at 13:26










          • $begingroup$
            Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
            $endgroup$
            – Silent
            Dec 2 '18 at 13:40










          • $begingroup$
            Yes, that is correct.
            $endgroup$
            – José Carlos Santos
            Dec 2 '18 at 14:25










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Silent
            Dec 2 '18 at 14:27


















          • $begingroup$
            thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
            $endgroup$
            – Silent
            Dec 2 '18 at 13:26










          • $begingroup$
            Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
            $endgroup$
            – Silent
            Dec 2 '18 at 13:40










          • $begingroup$
            Yes, that is correct.
            $endgroup$
            – José Carlos Santos
            Dec 2 '18 at 14:25










          • $begingroup$
            Thank you very much!
            $endgroup$
            – Silent
            Dec 2 '18 at 14:27
















          $begingroup$
          thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
          $endgroup$
          – Silent
          Dec 2 '18 at 13:26




          $begingroup$
          thank you very much for this answer. i was so silly to look from one side of $frac{c+d}{2}$ only.
          $endgroup$
          – Silent
          Dec 2 '18 at 13:26












          $begingroup$
          Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
          $endgroup$
          – Silent
          Dec 2 '18 at 13:40




          $begingroup$
          Please check my reasoning for why we can assume that there is an interval $(c,d)subset[a,b]$ such that $f(c)=0$ but $f(x)ne0$ for $c<x<d$: We know that the set ${xin[a,b]: f(x)=0}$ is closed, since $f$ continuous, hence the set at which function is nonzero is open. Since open set of $[a,b]$ are intersection of $[a,b]$ with at most countable open intervals, we can choose desired $(c,d)$ if we assume $f$ nonzero for some point in domain. Am I correct?
          $endgroup$
          – Silent
          Dec 2 '18 at 13:40












          $begingroup$
          Yes, that is correct.
          $endgroup$
          – José Carlos Santos
          Dec 2 '18 at 14:25




          $begingroup$
          Yes, that is correct.
          $endgroup$
          – José Carlos Santos
          Dec 2 '18 at 14:25












          $begingroup$
          Thank you very much!
          $endgroup$
          – Silent
          Dec 2 '18 at 14:27




          $begingroup$
          Thank you very much!
          $endgroup$
          – Silent
          Dec 2 '18 at 14:27


















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