Determining uniform convergence of a sequence of exponential functions
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I am given the sequence of functions $f_n(x)=e^{-nx^2}$ on $[-10,10]$. I must find the pointwise limit function $f(x)$ and decide whether convergence is uniform. If it is, I must find a $B_n$ such that $f_n(x) leq B_n$ for all $n$ and $B_n rightarrow 0$. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$
I determined the pointwise limit function to be $f(x)=begin{cases}
1 & x = 0 newline
0 & otherwise
end{cases}$
However, I am having trouble with the second step. I presume it is not uniformly convergent, since it will have a jump as $x rightarrow 0$. I do not know what $epsilon$ to work with here. I tried solving it for $x_n$ but I got something like $f_n(x_n) leq -epsilon$ which doesn't seem right.
real-analysis uniform-convergence
asked Dec 13 '18 at 17:18
hkj447hkj447
575
$endgroup$
|
show 1 more comment
$begingroup$
I am given the sequence of functions $f_n(x)=e^{-nx^2}$ on $[-10,10]$. I must find the pointwise limit function $f(x)$ and decide whether convergence is uniform. If it is, I must find a $B_n$ such that $f_n(x) leq B_n$ for all $n$ and $B_n rightarrow 0$. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$
I determined the pointwise limit function to be $f(x)=begin{cases}
1 & x = 0 newline
0 & otherwise
end{cases}$
However, I am having trouble with the second step. I presume it is not uniformly convergent, since it will have a jump as $x rightarrow 0$. I do not know what $epsilon$ to work with here. I tried solving it for $x_n$ but I got something like $f_n(x_n) leq -epsilon$ which doesn't seem right.
real-analysis uniform-convergence
asked Dec 13 '18 at 17:18
hkj447hkj447
575
$endgroup$
$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
1
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
1
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45
|
show 1 more comment
$begingroup$
I am given the sequence of functions $f_n(x)=e^{-nx^2}$ on $[-10,10]$. I must find the pointwise limit function $f(x)$ and decide whether convergence is uniform. If it is, I must find a $B_n$ such that $f_n(x) leq B_n$ for all $n$ and $B_n rightarrow 0$. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$
I determined the pointwise limit function to be $f(x)=begin{cases}
1 & x = 0 newline
0 & otherwise
end{cases}$
However, I am having trouble with the second step. I presume it is not uniformly convergent, since it will have a jump as $x rightarrow 0$. I do not know what $epsilon$ to work with here. I tried solving it for $x_n$ but I got something like $f_n(x_n) leq -epsilon$ which doesn't seem right.
real-analysis uniform-convergence
asked Dec 13 '18 at 17:18
hkj447hkj447
575
$endgroup$
I am given the sequence of functions $f_n(x)=e^{-nx^2}$ on $[-10,10]$. I must find the pointwise limit function $f(x)$ and decide whether convergence is uniform. If it is, I must find a $B_n$ such that $f_n(x) leq B_n$ for all $n$ and $B_n rightarrow 0$. If not, I must find an $epsilon > 0$ and a sequence $x_n$ such that $|f_n(x_n) - f(x_n)| geq epsilon$ for all $n$
I determined the pointwise limit function to be $f(x)=begin{cases}
1 & x = 0 newline
0 & otherwise
end{cases}$
However, I am having trouble with the second step. I presume it is not uniformly convergent, since it will have a jump as $x rightarrow 0$. I do not know what $epsilon$ to work with here. I tried solving it for $x_n$ but I got something like $f_n(x_n) leq -epsilon$ which doesn't seem right.
real-analysis uniform-convergence
real-analysis uniform-convergence
asked Dec 13 '18 at 17:18
hkj447hkj447
575
asked Dec 13 '18 at 17:18
hkj447hkj447
575
edited Dec 13 '18 at 17:31
hkj447
asked Dec 13 '18 at 17:18
hkj447hkj447
575
asked Dec 13 '18 at 17:18
hkj447hkj447
575
asked Dec 13 '18 at 17:18
hkj447hkj447
575
575
$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
1
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
1
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45
|
show 1 more comment
$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
1
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
1
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45
$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
1
1
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
1
1
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45
|
show 1 more comment
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$begingroup$
You can basically pick a value for $varepsilon$, e.g., $varepsilon=frac 12$. Observe that $f_n(0)=1>frac 12$ for all $n$ and $f_n(10)<frac 12$. Since $f_n$ is continuous, there must be some $x_nin(0,10)$ such that $f_n(x_n)=frac 12$. Now you should be able to solve for $x_n$.
$endgroup$
– cthl
Dec 13 '18 at 17:35
$begingroup$
So would any value for 0 < epsilon < 1 work?
$endgroup$
– hkj447
Dec 13 '18 at 17:41
1
$begingroup$
Strictly speaking you need that $varepsilon > f_1(10)$. Otherwise, the desired statement is only true for all but finitely many $n$. Alternatively, you can let $x_n$ be the minimum of $10$ and the solution of $f_n(y)=varepsilon$. But yes, besides technicalities, any $varepsilonin(0,1)$ will work.
$endgroup$
– cthl
Dec 13 '18 at 17:43
$begingroup$
I seem to run into an issue where I am taking the square root of a negative number when solving for Xn since I will be taking the ln of a number less than one
$endgroup$
– hkj447
Dec 13 '18 at 17:58
1
$begingroup$
But you also have another minus to make up for it: $e^{-nx^2}=frac 12$ $Rightarrow -nx^2=log(frac 12)$ $Rightarrow x^2=-frac 1nlog(frac 12)>0$.
$endgroup$
– cthl
Dec 13 '18 at 18:45