Inequality for a function 4

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Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that



$$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.



Is it correct that
$$u(t)leqfrac1sqrt N+frac1N^frac14$$
for every $tgeq 0$?
How could I prove that?




real-analysis analysis inequality




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    favorite












    Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that



    $$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.



    Is it correct that
    $$u(t)leqfrac1sqrt N+frac1N^frac14$$
    for every $tgeq 0$?
    How could I prove that?




    real-analysis analysis inequality




    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that



      $$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.



      Is it correct that
      $$u(t)leqfrac1sqrt N+frac1N^frac14$$
      for every $tgeq 0$?
      How could I prove that?




      real-analysis analysis inequality




      share|cite|improve this question














      Let $u:[0,+infty)tomathbb R^+$ be a bounded positive function such that



      $$u(t)leq int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14$$ for every $tgeq 0$, where $Ninmathbb N$.



      Is it correct that
      $$u(t)leqfrac1sqrt N+frac1N^frac14$$
      for every $tgeq 0$?
      How could I prove that?





      real-analysis analysis inequality





      share|cite|improve this question













      share|cite|improve this question




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      edited Aug 26 at 21:12









      zahbaz

      7,65521636




      7,65521636










      asked Aug 26 at 21:01









      user495333

      816




      816




















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          The inequality does not hold. As a counterexample, consider
          $$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
          Then one can show that $u$ satisfies the first inequality for $N = 1$:
          $$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
          and
          $$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
          so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$






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            The inequality does not hold. As a counterexample, consider
            $$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
            Then one can show that $u$ satisfies the first inequality for $N = 1$:
            $$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
            and
            $$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
            so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$






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              The inequality does not hold. As a counterexample, consider
              $$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
              Then one can show that $u$ satisfies the first inequality for $N = 1$:
              $$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
              and
              $$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
              so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$






              share|cite|improve this answer






















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                down vote










                up vote
                0
                down vote









                The inequality does not hold. As a counterexample, consider
                $$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
                Then one can show that $u$ satisfies the first inequality for $N = 1$:
                $$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
                and
                $$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
                so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$






                share|cite|improve this answer












                The inequality does not hold. As a counterexample, consider
                $$ u(t) = begincases1/2, &0leq t < 8 \ 3, &8leq t<9 \ 1/2, &9leq t<inftyendcases $$
                Then one can show that $u$ satisfies the first inequality for $N = 1$:
                $$ int_0^tleft(-frac1sqrt Nu(s)+frac1Nright)ds +frac1N^frac14 = 1 + int_0^t(1-u(s)),ds $$
                and
                $$1 + int_0^t(1-u(s)),ds = begincases1 + frac 1 2t, &0leq t < 8 \ 5 - 2(t-8), &8leq t<9 \ 3 + frac12(t-9), &9leq t<inftyendcases, $$
                so $u(t) leq 1 + int_0^t(1-u(s)), ds$. But $$u(8) = 3 > 2 = frac 1 sqrt N + frac 1 N^1/4.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Aug 26 at 21:42









                Trevor Norton

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