Proving $f_{s}(t) = frac{e^{-st}}{t}$ is uniformly convergent












1












$begingroup$


I am new to proving uniform convergence and am trying to show that $$f_{s}(t) = frac{e^{-st}}{t}$$ is uniformly convergent $forall$ $sgeq 0$. I am pretty sure that $tepsilon mathbb{R}$. I first thought of taking the derivative of the function to find the maximum since t is in an open set, but I remember that since I am showing that $left |f_{s}(t)right |<varepsilon$, and knowing that the value of $f_{s}(t)$ when $t<0$ (the value of the local maximum of $f_{s}(t)$ is also less than $0$) is negative, if one were to take the absolute value, then the local maximum would become a local minimum. Also, there would not be a maximum, since $lim _{srightarrow0} rightarrow+infty$. I know that the answer to this question is that $f_{s}(t)$ is uniformly convergent to $0$ as $trightarrow+infty$,$forall$ $sgeq 0$, but I am stuck at the point where I am at right now. Is there any theorem or something that I can use to help me with my proof?



Edit: This is in the context to use the Dirichlet test to show that an integrand composed of a product is uniformly convergent, so taking that in mind, should it be restricted to only the positive real numbers?






real-analysis uniform-convergence







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$endgroup$

















    1












    $begingroup$


    I am new to proving uniform convergence and am trying to show that $$f_{s}(t) = frac{e^{-st}}{t}$$ is uniformly convergent $forall$ $sgeq 0$. I am pretty sure that $tepsilon mathbb{R}$. I first thought of taking the derivative of the function to find the maximum since t is in an open set, but I remember that since I am showing that $left |f_{s}(t)right |<varepsilon$, and knowing that the value of $f_{s}(t)$ when $t<0$ (the value of the local maximum of $f_{s}(t)$ is also less than $0$) is negative, if one were to take the absolute value, then the local maximum would become a local minimum. Also, there would not be a maximum, since $lim _{srightarrow0} rightarrow+infty$. I know that the answer to this question is that $f_{s}(t)$ is uniformly convergent to $0$ as $trightarrow+infty$,$forall$ $sgeq 0$, but I am stuck at the point where I am at right now. Is there any theorem or something that I can use to help me with my proof?



    Edit: This is in the context to use the Dirichlet test to show that an integrand composed of a product is uniformly convergent, so taking that in mind, should it be restricted to only the positive real numbers?






    real-analysis uniform-convergence







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am new to proving uniform convergence and am trying to show that $$f_{s}(t) = frac{e^{-st}}{t}$$ is uniformly convergent $forall$ $sgeq 0$. I am pretty sure that $tepsilon mathbb{R}$. I first thought of taking the derivative of the function to find the maximum since t is in an open set, but I remember that since I am showing that $left |f_{s}(t)right |<varepsilon$, and knowing that the value of $f_{s}(t)$ when $t<0$ (the value of the local maximum of $f_{s}(t)$ is also less than $0$) is negative, if one were to take the absolute value, then the local maximum would become a local minimum. Also, there would not be a maximum, since $lim _{srightarrow0} rightarrow+infty$. I know that the answer to this question is that $f_{s}(t)$ is uniformly convergent to $0$ as $trightarrow+infty$,$forall$ $sgeq 0$, but I am stuck at the point where I am at right now. Is there any theorem or something that I can use to help me with my proof?



      Edit: This is in the context to use the Dirichlet test to show that an integrand composed of a product is uniformly convergent, so taking that in mind, should it be restricted to only the positive real numbers?






      real-analysis uniform-convergence







      share|cite|improve this question











      $endgroup$




      I am new to proving uniform convergence and am trying to show that $$f_{s}(t) = frac{e^{-st}}{t}$$ is uniformly convergent $forall$ $sgeq 0$. I am pretty sure that $tepsilon mathbb{R}$. I first thought of taking the derivative of the function to find the maximum since t is in an open set, but I remember that since I am showing that $left |f_{s}(t)right |<varepsilon$, and knowing that the value of $f_{s}(t)$ when $t<0$ (the value of the local maximum of $f_{s}(t)$ is also less than $0$) is negative, if one were to take the absolute value, then the local maximum would become a local minimum. Also, there would not be a maximum, since $lim _{srightarrow0} rightarrow+infty$. I know that the answer to this question is that $f_{s}(t)$ is uniformly convergent to $0$ as $trightarrow+infty$,$forall$ $sgeq 0$, but I am stuck at the point where I am at right now. Is there any theorem or something that I can use to help me with my proof?



      Edit: This is in the context to use the Dirichlet test to show that an integrand composed of a product is uniformly convergent, so taking that in mind, should it be restricted to only the positive real numbers?






      real-analysis uniform-convergence




      real-analysis uniform-convergence






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 29 '18 at 4:18









      RRL

      52.8k42573




      52.8k42573










      asked Dec 29 '18 at 1:02









      Eliot Eliot

      676




      676






















          1 Answer
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          $begingroup$

          For uniform convergence as $t to +infty$ you can consider the behavior when $t$ is positive. For all $s geqslant 0$ we have



          $$0 < frac{e^{-st}}{t} leqslant frac{1}{t}$$



          and it is now easy to conclude that convergence is uniform.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
            $endgroup$
            – Eliot
            Dec 29 '18 at 3:30










          • $begingroup$
            @Eliot: You're welcome.
            $endgroup$
            – RRL
            Dec 29 '18 at 3:43










          • $begingroup$
            Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:05












          • $begingroup$
            Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
            $endgroup$
            – RRL
            Dec 29 '18 at 4:14












          • $begingroup$
            Ok, that makes sense. Thank you again.
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:19











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          1 Answer
          1






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          active

          oldest

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          1












          $begingroup$

          For uniform convergence as $t to +infty$ you can consider the behavior when $t$ is positive. For all $s geqslant 0$ we have



          $$0 < frac{e^{-st}}{t} leqslant frac{1}{t}$$



          and it is now easy to conclude that convergence is uniform.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
            $endgroup$
            – Eliot
            Dec 29 '18 at 3:30










          • $begingroup$
            @Eliot: You're welcome.
            $endgroup$
            – RRL
            Dec 29 '18 at 3:43










          • $begingroup$
            Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:05












          • $begingroup$
            Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
            $endgroup$
            – RRL
            Dec 29 '18 at 4:14












          • $begingroup$
            Ok, that makes sense. Thank you again.
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:19
















          1












          $begingroup$

          For uniform convergence as $t to +infty$ you can consider the behavior when $t$ is positive. For all $s geqslant 0$ we have



          $$0 < frac{e^{-st}}{t} leqslant frac{1}{t}$$



          and it is now easy to conclude that convergence is uniform.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
            $endgroup$
            – Eliot
            Dec 29 '18 at 3:30










          • $begingroup$
            @Eliot: You're welcome.
            $endgroup$
            – RRL
            Dec 29 '18 at 3:43










          • $begingroup$
            Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:05












          • $begingroup$
            Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
            $endgroup$
            – RRL
            Dec 29 '18 at 4:14












          • $begingroup$
            Ok, that makes sense. Thank you again.
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:19














          1












          1








          1





          $begingroup$

          For uniform convergence as $t to +infty$ you can consider the behavior when $t$ is positive. For all $s geqslant 0$ we have



          $$0 < frac{e^{-st}}{t} leqslant frac{1}{t}$$



          and it is now easy to conclude that convergence is uniform.






          share|cite|improve this answer









          $endgroup$



          For uniform convergence as $t to +infty$ you can consider the behavior when $t$ is positive. For all $s geqslant 0$ we have



          $$0 < frac{e^{-st}}{t} leqslant frac{1}{t}$$



          and it is now easy to conclude that convergence is uniform.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 2:20









          RRLRRL

          52.8k42573




          52.8k42573












          • $begingroup$
            Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
            $endgroup$
            – Eliot
            Dec 29 '18 at 3:30










          • $begingroup$
            @Eliot: You're welcome.
            $endgroup$
            – RRL
            Dec 29 '18 at 3:43










          • $begingroup$
            Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:05












          • $begingroup$
            Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
            $endgroup$
            – RRL
            Dec 29 '18 at 4:14












          • $begingroup$
            Ok, that makes sense. Thank you again.
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:19


















          • $begingroup$
            Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
            $endgroup$
            – Eliot
            Dec 29 '18 at 3:30










          • $begingroup$
            @Eliot: You're welcome.
            $endgroup$
            – RRL
            Dec 29 '18 at 3:43










          • $begingroup$
            Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:05












          • $begingroup$
            Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
            $endgroup$
            – RRL
            Dec 29 '18 at 4:14












          • $begingroup$
            Ok, that makes sense. Thank you again.
            $endgroup$
            – Eliot
            Dec 29 '18 at 4:19
















          $begingroup$
          Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
          $endgroup$
          – Eliot
          Dec 29 '18 at 3:30




          $begingroup$
          Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
          $endgroup$
          – Eliot
          Dec 29 '18 at 3:30












          $begingroup$
          @Eliot: You're welcome.
          $endgroup$
          – RRL
          Dec 29 '18 at 3:43




          $begingroup$
          @Eliot: You're welcome.
          $endgroup$
          – RRL
          Dec 29 '18 at 3:43












          $begingroup$
          Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
          $endgroup$
          – Eliot
          Dec 29 '18 at 4:05






          $begingroup$
          Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
          $endgroup$
          – Eliot
          Dec 29 '18 at 4:05














          $begingroup$
          Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
          $endgroup$
          – RRL
          Dec 29 '18 at 4:14






          $begingroup$
          Here we are showing uniform convergence to $0$ as $t to infty$ for all values of the parameter $s geqslant 0$. It's uniform because $|f_s(t) - 0| < 1/t < epsilon$ for $t > T(epsilon) = 1/epsilon$ and $T(epsilon)$ does not depend on $s$.
          $endgroup$
          – RRL
          Dec 29 '18 at 4:14














          $begingroup$
          Ok, that makes sense. Thank you again.
          $endgroup$
          – Eliot
          Dec 29 '18 at 4:19




          $begingroup$
          Ok, that makes sense. Thank you again.
          $endgroup$
          – Eliot
          Dec 29 '18 at 4:19


















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