Mixing
Counterexample to Riemann sum limit
$begingroup$
For convergent Riemann sums of $f in C^1([0,1])$ there is the property:
$$tag{A}lim_{n to + infty} left[, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) , right] = int_0^1 f (x) dx$$
Proof using the mean value theorem goes like
$$, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) = f left( frac{n}{n+1} right) + sum_{k=0}^{n-1}f'(theta_k)left(frac{k}{n+1} - frac{k}{n} right)$$
and then shows the right side converges to $f(1) - int_0^1 xf'(x) dx = int_0^1f(x)dx.$ So it is very helpful that $f in C^1$.
If we only have $f in C[0,1]$, then equation (A) may not be true, and this was shown by Daniel Fischer using Banach-Steinhaus theorem in this answer. Here a related statement was considered:
$$tag{B}lim_{ntoinfty}left[, sum_{k=0}^{n-1} fleft(frac{k}{n}right) - nint_0^1 f(x),dx, right] = frac{f(0) - f(1)}{2}$$
My question: Can someone please show me a concrete counterexample, i.e., a continuous but not continuously differentiable function where (A) or (B) is false?
real-analysis examples-counterexamples riemann-integration
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
$endgroup$
add a comment |
$begingroup$
For convergent Riemann sums of $f in C^1([0,1])$ there is the property:
$$tag{A}lim_{n to + infty} left[, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) , right] = int_0^1 f (x) dx$$
Proof using the mean value theorem goes like
$$, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) = f left( frac{n}{n+1} right) + sum_{k=0}^{n-1}f'(theta_k)left(frac{k}{n+1} - frac{k}{n} right)$$
and then shows the right side converges to $f(1) - int_0^1 xf'(x) dx = int_0^1f(x)dx.$ So it is very helpful that $f in C^1$.
If we only have $f in C[0,1]$, then equation (A) may not be true, and this was shown by Daniel Fischer using Banach-Steinhaus theorem in this answer. Here a related statement was considered:
$$tag{B}lim_{ntoinfty}left[, sum_{k=0}^{n-1} fleft(frac{k}{n}right) - nint_0^1 f(x),dx, right] = frac{f(0) - f(1)}{2}$$
My question: Can someone please show me a concrete counterexample, i.e., a continuous but not continuously differentiable function where (A) or (B) is false?
real-analysis examples-counterexamples riemann-integration
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
$endgroup$
$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20
add a comment |
$begingroup$
For convergent Riemann sums of $f in C^1([0,1])$ there is the property:
$$tag{A}lim_{n to + infty} left[, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) , right] = int_0^1 f (x) dx$$
Proof using the mean value theorem goes like
$$, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) = f left( frac{n}{n+1} right) + sum_{k=0}^{n-1}f'(theta_k)left(frac{k}{n+1} - frac{k}{n} right)$$
and then shows the right side converges to $f(1) - int_0^1 xf'(x) dx = int_0^1f(x)dx.$ So it is very helpful that $f in C^1$.
If we only have $f in C[0,1]$, then equation (A) may not be true, and this was shown by Daniel Fischer using Banach-Steinhaus theorem in this answer. Here a related statement was considered:
$$tag{B}lim_{ntoinfty}left[, sum_{k=0}^{n-1} fleft(frac{k}{n}right) - nint_0^1 f(x),dx, right] = frac{f(0) - f(1)}{2}$$
My question: Can someone please show me a concrete counterexample, i.e., a continuous but not continuously differentiable function where (A) or (B) is false?
real-analysis examples-counterexamples riemann-integration
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
$endgroup$
For convergent Riemann sums of $f in C^1([0,1])$ there is the property:
$$tag{A}lim_{n to + infty} left[, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) , right] = int_0^1 f (x) dx$$
Proof using the mean value theorem goes like
$$, sum_{k=0}^{n} f left( frac{k}{n+1} right) - sum_{k=0}^{n-1} f left( frac{k}{n} right) = f left( frac{n}{n+1} right) + sum_{k=0}^{n-1}f'(theta_k)left(frac{k}{n+1} - frac{k}{n} right)$$
and then shows the right side converges to $f(1) - int_0^1 xf'(x) dx = int_0^1f(x)dx.$ So it is very helpful that $f in C^1$.
If we only have $f in C[0,1]$, then equation (A) may not be true, and this was shown by Daniel Fischer using Banach-Steinhaus theorem in this answer. Here a related statement was considered:
$$tag{B}lim_{ntoinfty}left[, sum_{k=0}^{n-1} fleft(frac{k}{n}right) - nint_0^1 f(x),dx, right] = frac{f(0) - f(1)}{2}$$
My question: Can someone please show me a concrete counterexample, i.e., a continuous but not continuously differentiable function where (A) or (B) is false?
real-analysis examples-counterexamples riemann-integration
real-analysis examples-counterexamples riemann-integration
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
edited Jan 29 at 15:08
YuiTo Cheng
2,0592637
2,0592637
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
asked Dec 10 '18 at 21:53
WoodWorkerWoodWorker
470314
470314
$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20
add a comment |
$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20
$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider Weierstrass function
$$
f(x)=sum_{l=1}^infty 2^{-l/2}cos pi 2^l x.
$$
It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.
Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=1}^infty 2^{-l/2}sum_{k=0}^{2^m-1}cos pi 2^{l-m} k.
$$
It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that
$$
sum_{k=0}^{2^m-1}cos pi 2^{l-m} k=
left{
begin{array}{r,l}
2^m,&&m<l,\
0,&&mge l.
end{array}
right.
$$
Therefore
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=m+1}^infty 2^{-l/2}2^m=left(1+sqrt{2}right) 2^{m/2}to infty
$$
as $mtoinfty$.
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
$endgroup$
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Consider Weierstrass function
$$
f(x)=sum_{l=1}^infty 2^{-l/2}cos pi 2^l x.
$$
It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.
Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=1}^infty 2^{-l/2}sum_{k=0}^{2^m-1}cos pi 2^{l-m} k.
$$
It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that
$$
sum_{k=0}^{2^m-1}cos pi 2^{l-m} k=
left{
begin{array}{r,l}
2^m,&&m<l,\
0,&&mge l.
end{array}
right.
$$
Therefore
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=m+1}^infty 2^{-l/2}2^m=left(1+sqrt{2}right) 2^{m/2}to infty
$$
as $mtoinfty$.
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
$endgroup$
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
add a comment |
$begingroup$
Consider Weierstrass function
$$
f(x)=sum_{l=1}^infty 2^{-l/2}cos pi 2^l x.
$$
It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.
Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=1}^infty 2^{-l/2}sum_{k=0}^{2^m-1}cos pi 2^{l-m} k.
$$
It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that
$$
sum_{k=0}^{2^m-1}cos pi 2^{l-m} k=
left{
begin{array}{r,l}
2^m,&&m<l,\
0,&&mge l.
end{array}
right.
$$
Therefore
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=m+1}^infty 2^{-l/2}2^m=left(1+sqrt{2}right) 2^{m/2}to infty
$$
as $mtoinfty$.
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
$endgroup$
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
add a comment |
$begingroup$
Consider Weierstrass function
$$
f(x)=sum_{l=1}^infty 2^{-l/2}cos pi 2^l x.
$$
It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.
Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=1}^infty 2^{-l/2}sum_{k=0}^{2^m-1}cos pi 2^{l-m} k.
$$
It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that
$$
sum_{k=0}^{2^m-1}cos pi 2^{l-m} k=
left{
begin{array}{r,l}
2^m,&&m<l,\
0,&&mge l.
end{array}
right.
$$
Therefore
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=m+1}^infty 2^{-l/2}2^m=left(1+sqrt{2}right) 2^{m/2}to infty
$$
as $mtoinfty$.
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
$endgroup$
Consider Weierstrass function
$$
f(x)=sum_{l=1}^infty 2^{-l/2}cos pi 2^l x.
$$
It is nowhere differentiable but is Holder continuous with exponent $1/2$. Also it is an example of a lacunary Fourier series.
Let's show that the expression under the limit sign in (B) is not bounded. The integral in the lhs is zero. For $n=2^m$
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=1}^infty 2^{-l/2}sum_{k=0}^{2^m-1}cos pi 2^{l-m} k.
$$
It's straightforward to check, presenting cosine as a sum of imaginary exponents and evaluating the appearing geometric progression, that
$$
sum_{k=0}^{2^m-1}cos pi 2^{l-m} k=
left{
begin{array}{r,l}
2^m,&&m<l,\
0,&&mge l.
end{array}
right.
$$
Therefore
$$
sum_{k=0}^{2^m-1}fleft(frac{k}{2^m}right)=
sum_{l=m+1}^infty 2^{-l/2}2^m=left(1+sqrt{2}right) 2^{m/2}to infty
$$
as $mtoinfty$.
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
answered Jan 26 at 9:21
AndrewAndrew
9,57211946
9,57211946
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
add a comment |
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
$begingroup$
Thank you, this stumped me?
$endgroup$
– WoodWorker
Jan 31 at 1:38
add a comment |
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$begingroup$
It is just an idea. Try to consider a real continuous function of unbounded variation on $[0,1]$, The right-hand side of (A) would be finite, however the left-hand side would be infinite.
$endgroup$
– Pavel R.
Dec 11 '18 at 0:46
$begingroup$
Banach Steinhaus (and the uniform boundedness principle, Baire category theorem, etc.) are constructive and can be unwound to provide counterexamples. Daniel Fischer's argument uses the fact that there are $f_n$ with $|T_n(f_n)|toinfty.$ Banach-Steinhaus is a recipe that computes some coefficients $c_n$ such that the linear combination $f=sum c_nf_n$ is uniformly convergent and satisfies $sup_n|T_n(f)|=infty.$
$endgroup$
– Dap
Jan 24 at 15:20