Show that the function $f$ is strictly bounded by an arbitrary number $A$












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$begingroup$



Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
$$f(a)<A$$
$$f(x)+f'(x)<A$$
Show that $f(x)<A$ for all $x in (a,b)$.




My attempt: Generate a function $h(x)$ such that
$$h(x)=e^xf(x)$$
$$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
$$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
$$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
$$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.






real-analysis derivatives







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    2












    $begingroup$



    Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
    $$f(a)<A$$
    $$f(x)+f'(x)<A$$
    Show that $f(x)<A$ for all $x in (a,b)$.




    My attempt: Generate a function $h(x)$ such that
    $$h(x)=e^xf(x)$$
    $$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
    So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
    $$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
    $$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
    $$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
    It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.






    real-analysis derivatives







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
      $$f(a)<A$$
      $$f(x)+f'(x)<A$$
      Show that $f(x)<A$ for all $x in (a,b)$.




      My attempt: Generate a function $h(x)$ such that
      $$h(x)=e^xf(x)$$
      $$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
      So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
      $$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
      $$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
      $$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
      It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.






      real-analysis derivatives







      share|cite|improve this question











      $endgroup$





      Suppose the function $f(x)$ is differentiable on $[a,b]$ and there exists an arbitrary number $A$ such that for all $x in (a,b)$
      $$f(a)<A$$
      $$f(x)+f'(x)<A$$
      Show that $f(x)<A$ for all $x in (a,b)$.




      My attempt: Generate a function $h(x)$ such that
      $$h(x)=e^xf(x)$$
      $$Longrightarrow h'(x)=[e^xf(x)]'=e^x[f(x)+f'(x)]<e^xA$$
      So $h'(x)$ is strictly bounded by $e^xA$. By MVT, there exists two points $c,d in (a,b)$ such that
      $$h(c)-h(d)=h'(frac{c+d}{2})(c-d)$$
      $$e^cf(c)<Ae^{frac{c+d}{2}}(c-d)+Ae^d$$
      $$f(c)<Ae^{frac{d-c}{2}}(c-d)+Ae^{d-c}=Ae^{d-c}[e^{frac{1}{2}}(c-d)+1]$$
      It seems my proof is wrong and get discontinued when I focus on simplifying for $f(c)$ but the question requires that $f(x)$ is strictly bounded by $A$ for the interval $I=(a,b)$.






      real-analysis derivatives




      real-analysis derivatives






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      edited Dec 20 '18 at 9:00







      weilam06

















      asked Dec 20 '18 at 7:11









      weilam06weilam06

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          $begingroup$

          I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
          Then
          $$h(c)-h(a)=int_a^c h'(t),dt
          <Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
          Then
          $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
          Then $f(c)<A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            my only worry here is given our definitions in the post $h$ might be defined at $a$?
            $endgroup$
            – user136920
            Dec 20 '18 at 7:42













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          $begingroup$

          I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
          Then
          $$h(c)-h(a)=int_a^c h'(t),dt
          <Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
          Then
          $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
          Then $f(c)<A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            my only worry here is given our definitions in the post $h$ might be defined at $a$?
            $endgroup$
            – user136920
            Dec 20 '18 at 7:42


















          1












          $begingroup$

          I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
          Then
          $$h(c)-h(a)=int_a^c h'(t),dt
          <Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
          Then
          $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
          Then $f(c)<A$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            my only worry here is given our definitions in the post $h$ might be defined at $a$?
            $endgroup$
            – user136920
            Dec 20 '18 at 7:42
















          1












          1








          1





          $begingroup$

          I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
          Then
          $$h(c)-h(a)=int_a^c h'(t),dt
          <Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
          Then
          $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
          Then $f(c)<A$.






          share|cite|improve this answer









          $endgroup$



          I don't see what you are bringing $c$ and $d$ into it. Let $cin(a,b)$.
          Then
          $$h(c)-h(a)=int_a^c h'(t),dt
          <Aint_{a}^c e^t,dt=Ae^c-Ae^a.$$
          Then
          $$h(c)<Ae^c-Ae^a+h(a)<Ae^c.$$
          Then $f(c)<A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 20 '18 at 7:21









          Lord Shark the UnknownLord Shark the Unknown

          105k1160133




          105k1160133












          • $begingroup$
            my only worry here is given our definitions in the post $h$ might be defined at $a$?
            $endgroup$
            – user136920
            Dec 20 '18 at 7:42




















          • $begingroup$
            my only worry here is given our definitions in the post $h$ might be defined at $a$?
            $endgroup$
            – user136920
            Dec 20 '18 at 7:42


















          $begingroup$
          my only worry here is given our definitions in the post $h$ might be defined at $a$?
          $endgroup$
          – user136920
          Dec 20 '18 at 7:42






          $begingroup$
          my only worry here is given our definitions in the post $h$ might be defined at $a$?
          $endgroup$
          – user136920
          Dec 20 '18 at 7:42




















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