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Proving $int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
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I came by the following theorem on a book that did not provide any proof.
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Since it is admitted that each $f_n$ is integrable I guess that each function could be integrated in the sum. However I fail to see a proof of the theorem.
Question:
How should I prove the claim?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
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up vote
1
down vote
favorite
I came by the following theorem on a book that did not provide any proof.
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Since it is admitted that each $f_n$ is integrable I guess that each function could be integrated in the sum. However I fail to see a proof of the theorem.
Question:
How should I prove the claim?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I came by the following theorem on a book that did not provide any proof.
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Since it is admitted that each $f_n$ is integrable I guess that each function could be integrated in the sum. However I fail to see a proof of the theorem.
Question:
How should I prove the claim?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
I came by the following theorem on a book that did not provide any proof.
Theorem: Suppose that $sum_limitsn=k^inftyf_n$ converges uniformly to $F$ on $S=[a,b]$. Assume that $F$ and $f_n::,ngeqslant k$, are integrable on $[a,b]$. Then:
$int_limitsa^bF(x)dx=sum_limitsn=k^inftyint_limitsa^bf_n(x)dx$
Since it is admitted that each $f_n$ is integrable I guess that each function could be integrated in the sum. However I fail to see a proof of the theorem.
Question:
How should I prove the claim?
Thanks in advance!
calculus real-analysis sequences-and-series
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
asked Jul 20 at 12:50
Pedro Gomes
1,3192618
1,3192618
Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58
add a comment |Â
Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58
Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58
add a comment |Â
1 Answer
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For $K ge k$, define $F_K: [a,b] to mathbb R$ by $$F_K(x) = sum^K_n=k f_n(x), ,,,, x in [a,b].$$ We're assuming $F_Kto F$ uniformly. Then for any $epsilon > 0$, there is $K^* in mathbb N$ such that $K > K^*$ ensures that $$lvert F(x) - F_K(x) rvert < epsilon / (b-a), ,,,, text for all x in [a,b].$$ Now for $K > K^*$, beginalign* left lvert int^b_a F(x) dx - int^b_a F_K(x) dx right rvert &= left lvert int^b_a [F(x) - F_K(x)] dx right rvert \
&le int^b_a lvert F(x) - F_K(x) rvert dx \
& le int^b_a fracepsilonb-a dx = epsilon.
endalign* This shows that $$lim_Ktoinfty int^b_a F_K(x) dx = int^b_a F(x) dx.$$ But by linearity of the integral, $$int^b_a F_K(x) dx = sum^K_n=k int^b_af_n(x) dx,$$ so $$lim_Ktoinfty sum^K_n=kint^b_a f_n(x) dx = int^b_a F(x) dx.$$ Or in other words, $$sum^infty_n=k int^b_a f_n(x) dx = int^b_a F(x) dx.$$
answered Jul 20 at 13:37
User8128
10.2k1522
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For $K ge k$, define $F_K: [a,b] to mathbb R$ by $$F_K(x) = sum^K_n=k f_n(x), ,,,, x in [a,b].$$ We're assuming $F_Kto F$ uniformly. Then for any $epsilon > 0$, there is $K^* in mathbb N$ such that $K > K^*$ ensures that $$lvert F(x) - F_K(x) rvert < epsilon / (b-a), ,,,, text for all x in [a,b].$$ Now for $K > K^*$, beginalign* left lvert int^b_a F(x) dx - int^b_a F_K(x) dx right rvert &= left lvert int^b_a [F(x) - F_K(x)] dx right rvert \
&le int^b_a lvert F(x) - F_K(x) rvert dx \
& le int^b_a fracepsilonb-a dx = epsilon.
endalign* This shows that $$lim_Ktoinfty int^b_a F_K(x) dx = int^b_a F(x) dx.$$ But by linearity of the integral, $$int^b_a F_K(x) dx = sum^K_n=k int^b_af_n(x) dx,$$ so $$lim_Ktoinfty sum^K_n=kint^b_a f_n(x) dx = int^b_a F(x) dx.$$ Or in other words, $$sum^infty_n=k int^b_a f_n(x) dx = int^b_a F(x) dx.$$
answered Jul 20 at 13:37
User8128
10.2k1522
add a comment |Â
up vote
3
down vote
accepted
For $K ge k$, define $F_K: [a,b] to mathbb R$ by $$F_K(x) = sum^K_n=k f_n(x), ,,,, x in [a,b].$$ We're assuming $F_Kto F$ uniformly. Then for any $epsilon > 0$, there is $K^* in mathbb N$ such that $K > K^*$ ensures that $$lvert F(x) - F_K(x) rvert < epsilon / (b-a), ,,,, text for all x in [a,b].$$ Now for $K > K^*$, beginalign* left lvert int^b_a F(x) dx - int^b_a F_K(x) dx right rvert &= left lvert int^b_a [F(x) - F_K(x)] dx right rvert \
&le int^b_a lvert F(x) - F_K(x) rvert dx \
& le int^b_a fracepsilonb-a dx = epsilon.
endalign* This shows that $$lim_Ktoinfty int^b_a F_K(x) dx = int^b_a F(x) dx.$$ But by linearity of the integral, $$int^b_a F_K(x) dx = sum^K_n=k int^b_af_n(x) dx,$$ so $$lim_Ktoinfty sum^K_n=kint^b_a f_n(x) dx = int^b_a F(x) dx.$$ Or in other words, $$sum^infty_n=k int^b_a f_n(x) dx = int^b_a F(x) dx.$$
answered Jul 20 at 13:37
User8128
10.2k1522
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For $K ge k$, define $F_K: [a,b] to mathbb R$ by $$F_K(x) = sum^K_n=k f_n(x), ,,,, x in [a,b].$$ We're assuming $F_Kto F$ uniformly. Then for any $epsilon > 0$, there is $K^* in mathbb N$ such that $K > K^*$ ensures that $$lvert F(x) - F_K(x) rvert < epsilon / (b-a), ,,,, text for all x in [a,b].$$ Now for $K > K^*$, beginalign* left lvert int^b_a F(x) dx - int^b_a F_K(x) dx right rvert &= left lvert int^b_a [F(x) - F_K(x)] dx right rvert \
&le int^b_a lvert F(x) - F_K(x) rvert dx \
& le int^b_a fracepsilonb-a dx = epsilon.
endalign* This shows that $$lim_Ktoinfty int^b_a F_K(x) dx = int^b_a F(x) dx.$$ But by linearity of the integral, $$int^b_a F_K(x) dx = sum^K_n=k int^b_af_n(x) dx,$$ so $$lim_Ktoinfty sum^K_n=kint^b_a f_n(x) dx = int^b_a F(x) dx.$$ Or in other words, $$sum^infty_n=k int^b_a f_n(x) dx = int^b_a F(x) dx.$$
answered Jul 20 at 13:37
User8128
10.2k1522
For $K ge k$, define $F_K: [a,b] to mathbb R$ by $$F_K(x) = sum^K_n=k f_n(x), ,,,, x in [a,b].$$ We're assuming $F_Kto F$ uniformly. Then for any $epsilon > 0$, there is $K^* in mathbb N$ such that $K > K^*$ ensures that $$lvert F(x) - F_K(x) rvert < epsilon / (b-a), ,,,, text for all x in [a,b].$$ Now for $K > K^*$, beginalign* left lvert int^b_a F(x) dx - int^b_a F_K(x) dx right rvert &= left lvert int^b_a [F(x) - F_K(x)] dx right rvert \
&le int^b_a lvert F(x) - F_K(x) rvert dx \
& le int^b_a fracepsilonb-a dx = epsilon.
endalign* This shows that $$lim_Ktoinfty int^b_a F_K(x) dx = int^b_a F(x) dx.$$ But by linearity of the integral, $$int^b_a F_K(x) dx = sum^K_n=k int^b_af_n(x) dx,$$ so $$lim_Ktoinfty sum^K_n=kint^b_a f_n(x) dx = int^b_a F(x) dx.$$ Or in other words, $$sum^infty_n=k int^b_a f_n(x) dx = int^b_a F(x) dx.$$
answered Jul 20 at 13:37
User8128
10.2k1522
answered Jul 20 at 13:37
User8128
10.2k1522
answered Jul 20 at 13:37
User8128
10.2k1522
answered Jul 20 at 13:37
User8128
10.2k1522
10.2k1522
add a comment |Â
add a comment |Â
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Which book are you referring to?
– uniquesolution
Jul 20 at 12:54
@uniquesolution Introduction to Real Analysis by William Trench.
– Pedro Gomes
Jul 20 at 12:58