Is this function of bounded variation?
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Consider Riemann's function defined on $mathbb{R}$,
$$ R(x) = sum_{n=1}^infty frac{sin n^2 x}{n^2} . $$
If you graph it, you can see that it shows a lot of zigzags.
Hence, the question is, is this function of bounded variation in $(0,2 pi )$?
real-analysis analysis fourier-analysis fourier-series
asked Nov 18 at 2:11
pie
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up vote
4
down vote
favorite
Consider Riemann's function defined on $mathbb{R}$,
$$ R(x) = sum_{n=1}^infty frac{sin n^2 x}{n^2} . $$
If you graph it, you can see that it shows a lot of zigzags.
Hence, the question is, is this function of bounded variation in $(0,2 pi )$?
real-analysis analysis fourier-analysis fourier-series
asked Nov 18 at 2:11
pie
685
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider Riemann's function defined on $mathbb{R}$,
$$ R(x) = sum_{n=1}^infty frac{sin n^2 x}{n^2} . $$
If you graph it, you can see that it shows a lot of zigzags.
Hence, the question is, is this function of bounded variation in $(0,2 pi )$?
real-analysis analysis fourier-analysis fourier-series
asked Nov 18 at 2:11
pie
685
Consider Riemann's function defined on $mathbb{R}$,
$$ R(x) = sum_{n=1}^infty frac{sin n^2 x}{n^2} . $$
If you graph it, you can see that it shows a lot of zigzags.
Hence, the question is, is this function of bounded variation in $(0,2 pi )$?
real-analysis analysis fourier-analysis fourier-series
real-analysis analysis fourier-analysis fourier-series
asked Nov 18 at 2:11
pie
685
asked Nov 18 at 2:11
pie
685
asked Nov 18 at 2:11
pie
685
asked Nov 18 at 2:11
pie
685
asked Nov 18 at 2:11
pie
685
685
add a comment |
add a comment |
2 Answers
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3
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Remember that functions of bounded variation are differentiable almost everywhere. See here for example that your $R(x)$ is differentiable nowhere. Thus by contraposition, it is not of bounded variation.
answered Nov 18 at 4:23
Zachary Selk
501211
add a comment |
up vote
1
down vote
I am just sharing my opinion. Please don't take it as an answer,
Differentiate $R(x)$ and take absolute value we get,
$$|R'(x)|lesum_{n=1}^infty |cos (n^2x)|$$
Is the RHS is bounded above by some quantity? If not then it is not of Bounded Variation.
Note: $|cos x|le 1,forall xinmathbb{R}$
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Remember that functions of bounded variation are differentiable almost everywhere. See here for example that your $R(x)$ is differentiable nowhere. Thus by contraposition, it is not of bounded variation.
answered Nov 18 at 4:23
Zachary Selk
501211
add a comment |
up vote
3
down vote
Remember that functions of bounded variation are differentiable almost everywhere. See here for example that your $R(x)$ is differentiable nowhere. Thus by contraposition, it is not of bounded variation.
answered Nov 18 at 4:23
Zachary Selk
501211
add a comment |
up vote
3
down vote
up vote
3
down vote
Remember that functions of bounded variation are differentiable almost everywhere. See here for example that your $R(x)$ is differentiable nowhere. Thus by contraposition, it is not of bounded variation.
answered Nov 18 at 4:23
Zachary Selk
501211
Remember that functions of bounded variation are differentiable almost everywhere. See here for example that your $R(x)$ is differentiable nowhere. Thus by contraposition, it is not of bounded variation.
answered Nov 18 at 4:23
Zachary Selk
501211
answered Nov 18 at 4:23
Zachary Selk
501211
answered Nov 18 at 4:23
Zachary Selk
501211
answered Nov 18 at 4:23
Zachary Selk
501211
501211
add a comment |
add a comment |
up vote
1
down vote
I am just sharing my opinion. Please don't take it as an answer,
Differentiate $R(x)$ and take absolute value we get,
$$|R'(x)|lesum_{n=1}^infty |cos (n^2x)|$$
Is the RHS is bounded above by some quantity? If not then it is not of Bounded Variation.
Note: $|cos x|le 1,forall xinmathbb{R}$
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
add a comment |
up vote
1
down vote
I am just sharing my opinion. Please don't take it as an answer,
Differentiate $R(x)$ and take absolute value we get,
$$|R'(x)|lesum_{n=1}^infty |cos (n^2x)|$$
Is the RHS is bounded above by some quantity? If not then it is not of Bounded Variation.
Note: $|cos x|le 1,forall xinmathbb{R}$
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
add a comment |
up vote
1
down vote
up vote
1
down vote
I am just sharing my opinion. Please don't take it as an answer,
Differentiate $R(x)$ and take absolute value we get,
$$|R'(x)|lesum_{n=1}^infty |cos (n^2x)|$$
Is the RHS is bounded above by some quantity? If not then it is not of Bounded Variation.
Note: $|cos x|le 1,forall xinmathbb{R}$
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
I am just sharing my opinion. Please don't take it as an answer,
Differentiate $R(x)$ and take absolute value we get,
$$|R'(x)|lesum_{n=1}^infty |cos (n^2x)|$$
Is the RHS is bounded above by some quantity? If not then it is not of Bounded Variation.
Note: $|cos x|le 1,forall xinmathbb{R}$
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
answered Nov 18 at 3:33
Sujit Bhattacharyya
905316
905316
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
add a comment |
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
@Zachary Selk Thanks for pointing out.. I am working on it. Just guessing, will Fubini's theorem work here. If I was unable to figure it out I'll remove the answer.
– Sujit Bhattacharyya
Nov 18 at 3:43
add a comment |
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