Let $f:[0,a]rightarrow[0,infty)$ be a continuous function such that $f(t)leq e^{int_{0}^{t}f(s)ds}-1$ for all...












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Let $f:[0,a]rightarrow[0,infty)$ be a continuous function such that $$f(t)leq e^{int_{0}^{t}f(s)ds}-1$$ for all $tin[0,a]$. Prove that $fequiv0$.



I have thought like this: Assume $F(t)=int_{0}^{t}f(s)dsimplies F'(t)=f(t)$ . Then $F'(t)leq e^{F(t)}-1$. Now how can proceed.






real-analysis inequality







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    $begingroup$


    Let $f:[0,a]rightarrow[0,infty)$ be a continuous function such that $$f(t)leq e^{int_{0}^{t}f(s)ds}-1$$ for all $tin[0,a]$. Prove that $fequiv0$.



    I have thought like this: Assume $F(t)=int_{0}^{t}f(s)dsimplies F'(t)=f(t)$ . Then $F'(t)leq e^{F(t)}-1$. Now how can proceed.






    real-analysis inequality







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $f:[0,a]rightarrow[0,infty)$ be a continuous function such that $$f(t)leq e^{int_{0}^{t}f(s)ds}-1$$ for all $tin[0,a]$. Prove that $fequiv0$.



      I have thought like this: Assume $F(t)=int_{0}^{t}f(s)dsimplies F'(t)=f(t)$ . Then $F'(t)leq e^{F(t)}-1$. Now how can proceed.






      real-analysis inequality







      share|cite|improve this question











      $endgroup$




      Let $f:[0,a]rightarrow[0,infty)$ be a continuous function such that $$f(t)leq e^{int_{0}^{t}f(s)ds}-1$$ for all $tin[0,a]$. Prove that $fequiv0$.



      I have thought like this: Assume $F(t)=int_{0}^{t}f(s)dsimplies F'(t)=f(t)$ . Then $F'(t)leq e^{F(t)}-1$. Now how can proceed.






      real-analysis inequality




      real-analysis inequality






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      share|cite|improve this question













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      edited yesterday









      YuiTo Cheng

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      2,43841037










      asked yesterday









      J.DoeJ.Doe

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          $begingroup$

          Since $$ F'(t) leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality
          $$e^{-F(t)-t} (F'(t)+1) leq e^{-t}$$
          $$ implies frac {d}{dt} e^{-F(t)-t} geq frac {d}{dt} e^{-t}$$
          $$implies e^{-F(t)-t} geq e^{-t} $$
          $$ implies e^{F(t)} leq 1$$
          $$implies F(t)leq 0$$
          But we know $F$ is non negative and hence $Fequiv 0$
          So $$ fequiv 0$$






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            $begingroup$

            Since $$ F'(t) leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality
            $$e^{-F(t)-t} (F'(t)+1) leq e^{-t}$$
            $$ implies frac {d}{dt} e^{-F(t)-t} geq frac {d}{dt} e^{-t}$$
            $$implies e^{-F(t)-t} geq e^{-t} $$
            $$ implies e^{F(t)} leq 1$$
            $$implies F(t)leq 0$$
            But we know $F$ is non negative and hence $Fequiv 0$
            So $$ fequiv 0$$






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              Since $$ F'(t) leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality
              $$e^{-F(t)-t} (F'(t)+1) leq e^{-t}$$
              $$ implies frac {d}{dt} e^{-F(t)-t} geq frac {d}{dt} e^{-t}$$
              $$implies e^{-F(t)-t} geq e^{-t} $$
              $$ implies e^{F(t)} leq 1$$
              $$implies F(t)leq 0$$
              But we know $F$ is non negative and hence $Fequiv 0$
              So $$ fequiv 0$$






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                Since $$ F'(t) leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality
                $$e^{-F(t)-t} (F'(t)+1) leq e^{-t}$$
                $$ implies frac {d}{dt} e^{-F(t)-t} geq frac {d}{dt} e^{-t}$$
                $$implies e^{-F(t)-t} geq e^{-t} $$
                $$ implies e^{F(t)} leq 1$$
                $$implies F(t)leq 0$$
                But we know $F$ is non negative and hence $Fequiv 0$
                So $$ fequiv 0$$






                share|cite|improve this answer









                $endgroup$



                Since $$ F'(t) leq e^{F(t)} -1$$ rearranging and multiplying we get the inequality
                $$e^{-F(t)-t} (F'(t)+1) leq e^{-t}$$
                $$ implies frac {d}{dt} e^{-F(t)-t} geq frac {d}{dt} e^{-t}$$
                $$implies e^{-F(t)-t} geq e^{-t} $$
                $$ implies e^{F(t)} leq 1$$
                $$implies F(t)leq 0$$
                But we know $F$ is non negative and hence $Fequiv 0$
                So $$ fequiv 0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered yesterday









                Ignorant MathematicianIgnorant Mathematician

                1,785114




                1,785114






























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