Proving $f_{s}(t) = frac{e^{-st}}{t}$ is uniformly convergent
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
edited Dec 29 '18 at 4:18
RRL
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
add a comment |
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
edited Dec 29 '18 at 4:18
RRL
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
add a comment |
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
edited Dec 29 '18 at 4:18
RRL
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
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
edited Dec 29 '18 at 4:18
RRL
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
edited Dec 29 '18 at 4:18
RRL
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
edited Dec 29 '18 at 4:18
RRL
49.4k42573
edited Dec 29 '18 at 4:18
RRL
49.4k42573
edited Dec 29 '18 at 4:18
RRL
49.4k42573
49.4k42573
asked Dec 29 '18 at 1:02
Eliot Eliot
676
asked Dec 29 '18 at 1:02
Eliot Eliot
676
asked Dec 29 '18 at 1:02
Eliot Eliot
676
676
add a comment |
add a comment |
1 Answer
1
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oldest
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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.
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
add a comment |
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.
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
add a comment |
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.
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
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.
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
answered Dec 29 '18 at 2:20
RRLRRL
49.4k42573
49.4k42573
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
add a comment |
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
Thank you I have not thought of that! It seems so obvious now but again I'm new to this material.
– Eliot
Dec 29 '18 at 3:30
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
@Eliot: You're welcome.
– RRL
Dec 29 '18 at 3:43
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
Actually isn't the relationship supposed to be independent of t? So wouldn't this prove point-wise convergence and not uniform convergence?
– Eliot
Dec 29 '18 at 4:05
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 $s$.
– RRL
Dec 29 '18 at 4:14
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 $s$.
– RRL
Dec 29 '18 at 4:14
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
Ok, that makes sense. Thank you again.
– Eliot
Dec 29 '18 at 4:19
add a comment |
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