Can a limit of an integral be moved inside the integral?












36












$begingroup$


After coming across this question: How to verify this limit, I have the following question:




When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?




For instance in the question, the OP is trying to determine that:



$$lim_{ntoinfty}int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}=1-rm{e}^{-1}$$



The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:



$$lim_{ntoinfty}{int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}}=int_{0}^{1}{lim_{ntoinfty}frac{dx}{(1+frac{x}{n})^{n}}}=int_{0}^{1}{frac{dx}{rm{e}^{x}}}=rm{e}^{0}-rm{e}^{-1}=1-rm{e}^{-1}$$



As required. So is this a valid technique, or is it just coincidental that this works?






calculus integration







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
    $endgroup$
    – user127.0.0.1
    Dec 8 '12 at 11:25








  • 12




    $begingroup$
    If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
    $endgroup$
    – user57159
    Dec 8 '12 at 11:31
















36












$begingroup$


After coming across this question: How to verify this limit, I have the following question:




When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?




For instance in the question, the OP is trying to determine that:



$$lim_{ntoinfty}int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}=1-rm{e}^{-1}$$



The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:



$$lim_{ntoinfty}{int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}}=int_{0}^{1}{lim_{ntoinfty}frac{dx}{(1+frac{x}{n})^{n}}}=int_{0}^{1}{frac{dx}{rm{e}^{x}}}=rm{e}^{0}-rm{e}^{-1}=1-rm{e}^{-1}$$



As required. So is this a valid technique, or is it just coincidental that this works?






calculus integration







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
    $endgroup$
    – user127.0.0.1
    Dec 8 '12 at 11:25








  • 12




    $begingroup$
    If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
    $endgroup$
    – user57159
    Dec 8 '12 at 11:31














36












36








36


39



$begingroup$


After coming across this question: How to verify this limit, I have the following question:




When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?




For instance in the question, the OP is trying to determine that:



$$lim_{ntoinfty}int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}=1-rm{e}^{-1}$$



The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:



$$lim_{ntoinfty}{int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}}=int_{0}^{1}{lim_{ntoinfty}frac{dx}{(1+frac{x}{n})^{n}}}=int_{0}^{1}{frac{dx}{rm{e}^{x}}}=rm{e}^{0}-rm{e}^{-1}=1-rm{e}^{-1}$$



As required. So is this a valid technique, or is it just coincidental that this works?






calculus integration







share|cite|improve this question











$endgroup$




After coming across this question: How to verify this limit, I have the following question:




When taking the limit of an integral, is it valid to move the limit inside the integral, providing the limit does not affect the limits of integration?




For instance in the question, the OP is trying to determine that:



$$lim_{ntoinfty}int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}=1-rm{e}^{-1}$$



The answers to the question involve evaluating the integral and then taking the limit to prove the result; but I was wondering if it would be valid to move the integral inside the limit, that is:



$$lim_{ntoinfty}{int_{0}^{1}{frac{dx}{(1+frac{x}{n})^{n}}}}=int_{0}^{1}{lim_{ntoinfty}frac{dx}{(1+frac{x}{n})^{n}}}=int_{0}^{1}{frac{dx}{rm{e}^{x}}}=rm{e}^{0}-rm{e}^{-1}=1-rm{e}^{-1}$$



As required. So is this a valid technique, or is it just coincidental that this works?






calculus integration




calculus integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 13 '17 at 12:21









Community♦

1




1










asked Dec 8 '12 at 11:21









Thomas RussellThomas Russell

7,89632552




7,89632552








  • 4




    $begingroup$
    It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
    $endgroup$
    – user127.0.0.1
    Dec 8 '12 at 11:25








  • 12




    $begingroup$
    If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
    $endgroup$
    – user57159
    Dec 8 '12 at 11:31














  • 4




    $begingroup$
    It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
    $endgroup$
    – user127.0.0.1
    Dec 8 '12 at 11:25








  • 12




    $begingroup$
    If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
    $endgroup$
    – user57159
    Dec 8 '12 at 11:31








4




4




$begingroup$
It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
$endgroup$
– user127.0.0.1
Dec 8 '12 at 11:25






$begingroup$
It depends... e.g. if $f_nlongrightarrow f$ uniform, you are allowed to do so.
$endgroup$
– user127.0.0.1
Dec 8 '12 at 11:25






12




12




$begingroup$
If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
$endgroup$
– user57159
Dec 8 '12 at 11:31




$begingroup$
If $: f_n longrightarrow f :$ uniformly and the interval is finite, then you are allowed to do so. $hspace{1.2 in}$
$endgroup$
– user57159
Dec 8 '12 at 11:31










2 Answers
2






active

oldest

votes


















30












$begingroup$

Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.



The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually, both DCT and MCT are equivalent.
    $endgroup$
    – EA304GT
    Oct 5 '16 at 0:08






  • 1




    $begingroup$
    @EA304GT 'both are equivalent' is redundant
    $endgroup$
    – mathworker21
    Jul 20 '18 at 12:33










  • $begingroup$
    If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
    $endgroup$
    – Hashimoto
    Feb 4 at 21:53



















22












$begingroup$

The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+frac{x}{n})^{-n}$, then
$$lim_{nto +infty}f_n(x)=e^{-x}=f(x)$$
We need to show that on $[0,1]$, $$left|f_n-f right|_{infty}to 0$$
But
$$left|f_n-f right|_{infty}=sup_{xin [0,1]}left|f_n(x)-f(x)right|=
sup_{xin [0,1]}left|(1+frac{x}{n})^{-n}-e^{-x}right|
$$
We need to determine the maximum of $g_n(x)=(1+frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.



$g_n(0)=0$, $g_n(1)=(1+frac1n)^{-n}-e^{-1}$ and
$$g_n^{prime}(x_0)=0Leftrightarrow (1+frac{x_0}{n})^{-n-1}=e^{-x_0}$$
Whenever the latter is true, $g_n(x_0)=frac{x_0}{n}e^{-x_0}$. Therefore,
$$left|f_n-f right|_{infty}=max_{xin [0,1]}left|g_n(x)right|=maxleft{g_n(0),g_n(1),g_n(x_0)right}=maxleft{0,(1+frac1n)^{-n}-e^{-1},frac{x_0}{n}e^{-x_0}right}to 0
$$
as $nto +infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
    $endgroup$
    – 2chromatic
    Sep 17 '18 at 11:11












  • $begingroup$
    Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
    $endgroup$
    – Crunchy
    Feb 28 at 2:56










  • $begingroup$
    The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
    $endgroup$
    – Gilles Castel
    Mar 13 at 14:13













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









30












$begingroup$

Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.



The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually, both DCT and MCT are equivalent.
    $endgroup$
    – EA304GT
    Oct 5 '16 at 0:08






  • 1




    $begingroup$
    @EA304GT 'both are equivalent' is redundant
    $endgroup$
    – mathworker21
    Jul 20 '18 at 12:33










  • $begingroup$
    If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
    $endgroup$
    – Hashimoto
    Feb 4 at 21:53
















30












$begingroup$

Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.



The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Actually, both DCT and MCT are equivalent.
    $endgroup$
    – EA304GT
    Oct 5 '16 at 0:08






  • 1




    $begingroup$
    @EA304GT 'both are equivalent' is redundant
    $endgroup$
    – mathworker21
    Jul 20 '18 at 12:33










  • $begingroup$
    If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
    $endgroup$
    – Hashimoto
    Feb 4 at 21:53














30












30








30





$begingroup$

Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.



The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.






share|cite|improve this answer









$endgroup$



Taking the limit inside the integral is not always allowed. There are several theorems that allow you to do so. The major ones being Lebesgue dominated convergence theorem and Monotone convergence theorem.



The uniform convergence mentioned in the comments is a special case of Dominated convergence theorem.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '12 at 11:31









CantorCantor

1,088821




1,088821












  • $begingroup$
    Actually, both DCT and MCT are equivalent.
    $endgroup$
    – EA304GT
    Oct 5 '16 at 0:08






  • 1




    $begingroup$
    @EA304GT 'both are equivalent' is redundant
    $endgroup$
    – mathworker21
    Jul 20 '18 at 12:33










  • $begingroup$
    If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
    $endgroup$
    – Hashimoto
    Feb 4 at 21:53


















  • $begingroup$
    Actually, both DCT and MCT are equivalent.
    $endgroup$
    – EA304GT
    Oct 5 '16 at 0:08






  • 1




    $begingroup$
    @EA304GT 'both are equivalent' is redundant
    $endgroup$
    – mathworker21
    Jul 20 '18 at 12:33










  • $begingroup$
    If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
    $endgroup$
    – Hashimoto
    Feb 4 at 21:53
















$begingroup$
Actually, both DCT and MCT are equivalent.
$endgroup$
– EA304GT
Oct 5 '16 at 0:08




$begingroup$
Actually, both DCT and MCT are equivalent.
$endgroup$
– EA304GT
Oct 5 '16 at 0:08




1




1




$begingroup$
@EA304GT 'both are equivalent' is redundant
$endgroup$
– mathworker21
Jul 20 '18 at 12:33




$begingroup$
@EA304GT 'both are equivalent' is redundant
$endgroup$
– mathworker21
Jul 20 '18 at 12:33












$begingroup$
If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
$endgroup$
– Hashimoto
Feb 4 at 21:53




$begingroup$
If you know that the limit function is integrable, uniform convergence is enough. DCT and MCT go a step further and ensure that the limit is also integrable.
$endgroup$
– Hashimoto
Feb 4 at 21:53











22












$begingroup$

The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+frac{x}{n})^{-n}$, then
$$lim_{nto +infty}f_n(x)=e^{-x}=f(x)$$
We need to show that on $[0,1]$, $$left|f_n-f right|_{infty}to 0$$
But
$$left|f_n-f right|_{infty}=sup_{xin [0,1]}left|f_n(x)-f(x)right|=
sup_{xin [0,1]}left|(1+frac{x}{n})^{-n}-e^{-x}right|
$$
We need to determine the maximum of $g_n(x)=(1+frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.



$g_n(0)=0$, $g_n(1)=(1+frac1n)^{-n}-e^{-1}$ and
$$g_n^{prime}(x_0)=0Leftrightarrow (1+frac{x_0}{n})^{-n-1}=e^{-x_0}$$
Whenever the latter is true, $g_n(x_0)=frac{x_0}{n}e^{-x_0}$. Therefore,
$$left|f_n-f right|_{infty}=max_{xin [0,1]}left|g_n(x)right|=maxleft{g_n(0),g_n(1),g_n(x_0)right}=maxleft{0,(1+frac1n)^{-n}-e^{-1},frac{x_0}{n}e^{-x_0}right}to 0
$$
as $nto +infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
    $endgroup$
    – 2chromatic
    Sep 17 '18 at 11:11












  • $begingroup$
    Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
    $endgroup$
    – Crunchy
    Feb 28 at 2:56










  • $begingroup$
    The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
    $endgroup$
    – Gilles Castel
    Mar 13 at 14:13


















22












$begingroup$

The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+frac{x}{n})^{-n}$, then
$$lim_{nto +infty}f_n(x)=e^{-x}=f(x)$$
We need to show that on $[0,1]$, $$left|f_n-f right|_{infty}to 0$$
But
$$left|f_n-f right|_{infty}=sup_{xin [0,1]}left|f_n(x)-f(x)right|=
sup_{xin [0,1]}left|(1+frac{x}{n})^{-n}-e^{-x}right|
$$
We need to determine the maximum of $g_n(x)=(1+frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.



$g_n(0)=0$, $g_n(1)=(1+frac1n)^{-n}-e^{-1}$ and
$$g_n^{prime}(x_0)=0Leftrightarrow (1+frac{x_0}{n})^{-n-1}=e^{-x_0}$$
Whenever the latter is true, $g_n(x_0)=frac{x_0}{n}e^{-x_0}$. Therefore,
$$left|f_n-f right|_{infty}=max_{xin [0,1]}left|g_n(x)right|=maxleft{g_n(0),g_n(1),g_n(x_0)right}=maxleft{0,(1+frac1n)^{-n}-e^{-1},frac{x_0}{n}e^{-x_0}right}to 0
$$
as $nto +infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
    $endgroup$
    – 2chromatic
    Sep 17 '18 at 11:11












  • $begingroup$
    Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
    $endgroup$
    – Crunchy
    Feb 28 at 2:56










  • $begingroup$
    The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
    $endgroup$
    – Gilles Castel
    Mar 13 at 14:13
















22












22








22





$begingroup$

The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+frac{x}{n})^{-n}$, then
$$lim_{nto +infty}f_n(x)=e^{-x}=f(x)$$
We need to show that on $[0,1]$, $$left|f_n-f right|_{infty}to 0$$
But
$$left|f_n-f right|_{infty}=sup_{xin [0,1]}left|f_n(x)-f(x)right|=
sup_{xin [0,1]}left|(1+frac{x}{n})^{-n}-e^{-x}right|
$$
We need to determine the maximum of $g_n(x)=(1+frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.



$g_n(0)=0$, $g_n(1)=(1+frac1n)^{-n}-e^{-1}$ and
$$g_n^{prime}(x_0)=0Leftrightarrow (1+frac{x_0}{n})^{-n-1}=e^{-x_0}$$
Whenever the latter is true, $g_n(x_0)=frac{x_0}{n}e^{-x_0}$. Therefore,
$$left|f_n-f right|_{infty}=max_{xin [0,1]}left|g_n(x)right|=maxleft{g_n(0),g_n(1),g_n(x_0)right}=maxleft{0,(1+frac1n)^{-n}-e^{-1},frac{x_0}{n}e^{-x_0}right}to 0
$$
as $nto +infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done






share|cite|improve this answer











$endgroup$



The limit can be moved inside the integral if the convergence of the integrand is uniform. In our case if $f_n(x)=(1+frac{x}{n})^{-n}$, then
$$lim_{nto +infty}f_n(x)=e^{-x}=f(x)$$
We need to show that on $[0,1]$, $$left|f_n-f right|_{infty}to 0$$
But
$$left|f_n-f right|_{infty}=sup_{xin [0,1]}left|f_n(x)-f(x)right|=
sup_{xin [0,1]}left|(1+frac{x}{n})^{-n}-e^{-x}right|
$$
We need to determine the maximum of $g_n(x)=(1+frac{x}{n})^{-n}-e^{-x}$ on $[0,1]$ or at least show it converges to $0$.



$g_n(0)=0$, $g_n(1)=(1+frac1n)^{-n}-e^{-1}$ and
$$g_n^{prime}(x_0)=0Leftrightarrow (1+frac{x_0}{n})^{-n-1}=e^{-x_0}$$
Whenever the latter is true, $g_n(x_0)=frac{x_0}{n}e^{-x_0}$. Therefore,
$$left|f_n-f right|_{infty}=max_{xin [0,1]}left|g_n(x)right|=maxleft{g_n(0),g_n(1),g_n(x_0)right}=maxleft{0,(1+frac1n)^{-n}-e^{-1},frac{x_0}{n}e^{-x_0}right}to 0
$$
as $nto +infty$. Uniform convergence on $[0,1]$ is proven and the limit-integral interchange can be done







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Oct 31 '18 at 14:12









AgentSmith

9615




9615










answered Dec 8 '12 at 11:50









NamelessNameless

10.4k12155




10.4k12155












  • $begingroup$
    Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
    $endgroup$
    – 2chromatic
    Sep 17 '18 at 11:11












  • $begingroup$
    Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
    $endgroup$
    – Crunchy
    Feb 28 at 2:56










  • $begingroup$
    The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
    $endgroup$
    – Gilles Castel
    Mar 13 at 14:13




















  • $begingroup$
    Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
    $endgroup$
    – 2chromatic
    Sep 17 '18 at 11:11












  • $begingroup$
    Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
    $endgroup$
    – Crunchy
    Feb 28 at 2:56










  • $begingroup$
    The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
    $endgroup$
    – Gilles Castel
    Mar 13 at 14:13


















$begingroup$
Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
$endgroup$
– 2chromatic
Sep 17 '18 at 11:11






$begingroup$
Why the $g_n(x_0)$ is not equal to $e^{-x_0}frac{x_0}{n}$ ?
$endgroup$
– 2chromatic
Sep 17 '18 at 11:11














$begingroup$
Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
$endgroup$
– Crunchy
Feb 28 at 2:56




$begingroup$
Hi @Nameless! I'm obviously a newbie here. So pardon me to ask this. Why is it sufficient to show that $| f_n-f|_{infty} rightarrow 0$ to show that $f_n$ converges uniformly? :)
$endgroup$
– Crunchy
Feb 28 at 2:56












$begingroup$
The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
$endgroup$
– Gilles Castel
Mar 13 at 14:13






$begingroup$
The expression $|f_n - f|_infty to 0$ is equivalent with $forall epsilon>0 :exists N in mathbb N: forall n > N: sup_{x in mathbb R} |f_n(x) - f(x)| < epsilon$. By the definition of supremum, we get that $|f_n(x) - f(x)| < epsilon$ for all $x in mathbb R$. In other words, for all $epsilon>0$, we find an $N$, independent of $x$ such that if $n>N$, $|f_n(x) - f(x) | < epsilon$ for all $x in mathbb R$, which is exactly the definition of uniform convergence.
$endgroup$
– Gilles Castel
Mar 13 at 14:13




















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