Mixing
Uniform convergence of $f_n(x) = n(x-1)e^{-nx}$
$begingroup$
I need to study the uniform convergence of
$f_n(x) = n(x-1)e^{-nx}$
on the interval $[0,+infty)$
I've shown that :
at $x =0$ $f_n(0)=-n xrightarrow{} -infty$
at $x =1$ $f_n(1)=0 xrightarrow{} 0$
for every $x neq 0$ we have that $f_n(x)xrightarrow{} 0$ for n $rightarrow$ + $infty$
It remains to prove the uniform convergence on the interval $[0,+infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x neq 0$. So in $[0,infty)$ there isn't uniform convergence?
Thanks in advance for any help.
real-analysis limits sequence-of-function
edited Jan 22 at 21:27
JustDroppedIn
2,153420
asked Jan 22 at 21:08
andrewandrew
698
$endgroup$
add a comment |
$begingroup$
I need to study the uniform convergence of
$f_n(x) = n(x-1)e^{-nx}$
on the interval $[0,+infty)$
I've shown that :
at $x =0$ $f_n(0)=-n xrightarrow{} -infty$
at $x =1$ $f_n(1)=0 xrightarrow{} 0$
for every $x neq 0$ we have that $f_n(x)xrightarrow{} 0$ for n $rightarrow$ + $infty$
It remains to prove the uniform convergence on the interval $[0,+infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x neq 0$. So in $[0,infty)$ there isn't uniform convergence?
Thanks in advance for any help.
real-analysis limits sequence-of-function
edited Jan 22 at 21:27
JustDroppedIn
2,153420
asked Jan 22 at 21:08
andrewandrew
698
$endgroup$
$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20
add a comment |
$begingroup$
I need to study the uniform convergence of
$f_n(x) = n(x-1)e^{-nx}$
on the interval $[0,+infty)$
I've shown that :
at $x =0$ $f_n(0)=-n xrightarrow{} -infty$
at $x =1$ $f_n(1)=0 xrightarrow{} 0$
for every $x neq 0$ we have that $f_n(x)xrightarrow{} 0$ for n $rightarrow$ + $infty$
It remains to prove the uniform convergence on the interval $[0,+infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x neq 0$. So in $[0,infty)$ there isn't uniform convergence?
Thanks in advance for any help.
real-analysis limits sequence-of-function
edited Jan 22 at 21:27
JustDroppedIn
2,153420
asked Jan 22 at 21:08
andrewandrew
698
$endgroup$
I need to study the uniform convergence of
$f_n(x) = n(x-1)e^{-nx}$
on the interval $[0,+infty)$
I've shown that :
at $x =0$ $f_n(0)=-n xrightarrow{} -infty$
at $x =1$ $f_n(1)=0 xrightarrow{} 0$
for every $x neq 0$ we have that $f_n(x)xrightarrow{} 0$ for n $rightarrow$ + $infty$
It remains to prove the uniform convergence on the interval $[0,+infty)$. But at $0$ there isn't pointwise convergence and the function $f(x)=0$ is defined for every $x neq 0$. So in $[0,infty)$ there isn't uniform convergence?
Thanks in advance for any help.
real-analysis limits sequence-of-function
real-analysis limits sequence-of-function
edited Jan 22 at 21:27
JustDroppedIn
2,153420
asked Jan 22 at 21:08
andrewandrew
698
edited Jan 22 at 21:27
JustDroppedIn
2,153420
asked Jan 22 at 21:08
andrewandrew
698
edited Jan 22 at 21:27
JustDroppedIn
2,153420
edited Jan 22 at 21:27
JustDroppedIn
2,153420
edited Jan 22 at 21:27
JustDroppedIn
2,153420
2,153420
asked Jan 22 at 21:08
andrewandrew
698
asked Jan 22 at 21:08
andrewandrew
698
asked Jan 22 at 21:08
andrewandrew
698
698
$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20
add a comment |
$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20
$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are right, it does not converge uniformly on $[0,infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $displaystyle{lim_{ntoinfty}sup_{x>0}|f_n(x)|=0}$. But $displaystyle{sup_{x>0}|f_n(x)|geqlim_{xto0^+}|f_n(x)|=n}$, hence the limit above is equal to $+infty$. Therefore the sequence does not converge uniformly on $(0,infty)$ either. Let's take some small $varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[varepsilon,infty)$. We have that $displaystyle{sup_{xgeqvarepsilon}|f_n(x)|=|f_n(varepsilon)|}$ (take $varepsilonleq1$, it is easy to verify it). Now it is $|f_n(varepsilon)|to0$, therefore the convergence is uniform on $[varepsilon,infty)$ for any $varepsilon>0$, but not on $[0,infty)$, nor $(0,infty)$.
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
add a comment |
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1 Answer
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$begingroup$
You are right, it does not converge uniformly on $[0,infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $displaystyle{lim_{ntoinfty}sup_{x>0}|f_n(x)|=0}$. But $displaystyle{sup_{x>0}|f_n(x)|geqlim_{xto0^+}|f_n(x)|=n}$, hence the limit above is equal to $+infty$. Therefore the sequence does not converge uniformly on $(0,infty)$ either. Let's take some small $varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[varepsilon,infty)$. We have that $displaystyle{sup_{xgeqvarepsilon}|f_n(x)|=|f_n(varepsilon)|}$ (take $varepsilonleq1$, it is easy to verify it). Now it is $|f_n(varepsilon)|to0$, therefore the convergence is uniform on $[varepsilon,infty)$ for any $varepsilon>0$, but not on $[0,infty)$, nor $(0,infty)$.
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
add a comment |
$begingroup$
You are right, it does not converge uniformly on $[0,infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $displaystyle{lim_{ntoinfty}sup_{x>0}|f_n(x)|=0}$. But $displaystyle{sup_{x>0}|f_n(x)|geqlim_{xto0^+}|f_n(x)|=n}$, hence the limit above is equal to $+infty$. Therefore the sequence does not converge uniformly on $(0,infty)$ either. Let's take some small $varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[varepsilon,infty)$. We have that $displaystyle{sup_{xgeqvarepsilon}|f_n(x)|=|f_n(varepsilon)|}$ (take $varepsilonleq1$, it is easy to verify it). Now it is $|f_n(varepsilon)|to0$, therefore the convergence is uniform on $[varepsilon,infty)$ for any $varepsilon>0$, but not on $[0,infty)$, nor $(0,infty)$.
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
$endgroup$
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
add a comment |
$begingroup$
You are right, it does not converge uniformly on $[0,infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $displaystyle{lim_{ntoinfty}sup_{x>0}|f_n(x)|=0}$. But $displaystyle{sup_{x>0}|f_n(x)|geqlim_{xto0^+}|f_n(x)|=n}$, hence the limit above is equal to $+infty$. Therefore the sequence does not converge uniformly on $(0,infty)$ either. Let's take some small $varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[varepsilon,infty)$. We have that $displaystyle{sup_{xgeqvarepsilon}|f_n(x)|=|f_n(varepsilon)|}$ (take $varepsilonleq1$, it is easy to verify it). Now it is $|f_n(varepsilon)|to0$, therefore the convergence is uniform on $[varepsilon,infty)$ for any $varepsilon>0$, but not on $[0,infty)$, nor $(0,infty)$.
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
$endgroup$
You are right, it does not converge uniformly on $[0,infty)$, since it does not even converge at point $0$. Observe that it is a sequence of continuous functions and if it converges uniformly, the limit is a continuous function. Now let's see if it can converge uniformly on $(0,infty)$. The only possible limit is the pointwise limit, which is the $0$ function. Now we want to show that $displaystyle{lim_{ntoinfty}sup_{x>0}|f_n(x)|=0}$. But $displaystyle{sup_{x>0}|f_n(x)|geqlim_{xto0^+}|f_n(x)|=n}$, hence the limit above is equal to $+infty$. Therefore the sequence does not converge uniformly on $(0,infty)$ either. Let's take some small $varepsilon>0$ and see if we can prove that $(f_n)$ converges uniformly on $[varepsilon,infty)$. We have that $displaystyle{sup_{xgeqvarepsilon}|f_n(x)|=|f_n(varepsilon)|}$ (take $varepsilonleq1$, it is easy to verify it). Now it is $|f_n(varepsilon)|to0$, therefore the convergence is uniform on $[varepsilon,infty)$ for any $varepsilon>0$, but not on $[0,infty)$, nor $(0,infty)$.
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
answered Jan 22 at 21:17
JustDroppedInJustDroppedIn
2,153420
2,153420
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
add a comment |
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
Can you explain in what way $sup_{x>0} |f_n(x)| geq |f_n(0)|=n$ is true? The maximal value is attained at $x=0$, so in the open interval any value $x>0$ will lead to a value $|f_n(x)|$ which is actually $< n$ regardless of how close I'm to $x=0$.
$endgroup$
– Diger
Jan 22 at 21:50
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
@Diger Take any sequence $(x_m)subset (0,infty)$ converging to $0$. Obviously $sup_{x>0}|f_n(x)|geq|f_n(x_m)|$ for any $m$. Take limits on this inequality as $mtoinfty$. The LHS is not affected by $m$. The RHS tends to $|f_n(0)|$ by the continuity of $f_n$.
$endgroup$
– JustDroppedIn
Jan 22 at 21:58
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
Yes it is true for every finite $m$, but I'm not sure if the limit can be taken. Feels a bit weird, since in that case there is no interval/value left on the left hand side of $x_m$ :-/
$endgroup$
– Diger
Jan 22 at 22:08
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
$begingroup$
@diger the limit can be taken. It is true for every finite m, so let m get as large as it wants. that's exactly what a limit is
$endgroup$
– JustDroppedIn
Jan 23 at 0:09
add a comment |
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$begingroup$
$f_n(0) = -n$ which is not bounded and hence not converging.
$endgroup$
– Will M.
Jan 22 at 21:12
$begingroup$
I'm pretty sure there is a mistake in the exercise and you should study uniform convergence in the open interval $(0,infty)$.
$endgroup$
– Mark
Jan 22 at 21:13
$begingroup$
indeed the next question of my exercise is studying the uniform convergence in $[1,infty]$. However i have to say that there isn't uniform convergence in $[0,+infty]$.
$endgroup$
– andrew
Jan 22 at 21:20