Mixing
Pointwise convergence and Dirichlet's theorem on Fourier series
Let $f(x)=xcos(alpha x)$ at $[-pi,pi]$ for $lambda in mathbb{R}$. Let $S$ be $f(x)$'s corresponding Fourier Series.
Prove or disprove: 1) if $lambda in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
2) If $lambda = m + dfrac{1}{2}$ for $m in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
Thoughts -
Since $f(x)$ is continuous (multiplication of continuous functions) in $[-pi,pi]$ we could use Dirichlet's theorem and just say that $S$ should converge to $f(x)$, it just seems a little bit naive. What am i missing ?
Thanks in advance!
fourier-series harmonic-analysis
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
add a comment |
Let $f(x)=xcos(alpha x)$ at $[-pi,pi]$ for $lambda in mathbb{R}$. Let $S$ be $f(x)$'s corresponding Fourier Series.
Prove or disprove: 1) if $lambda in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
2) If $lambda = m + dfrac{1}{2}$ for $m in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
Thoughts -
Since $f(x)$ is continuous (multiplication of continuous functions) in $[-pi,pi]$ we could use Dirichlet's theorem and just say that $S$ should converge to $f(x)$, it just seems a little bit naive. What am i missing ?
Thanks in advance!
fourier-series harmonic-analysis
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47
add a comment |
Let $f(x)=xcos(alpha x)$ at $[-pi,pi]$ for $lambda in mathbb{R}$. Let $S$ be $f(x)$'s corresponding Fourier Series.
Prove or disprove: 1) if $lambda in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
2) If $lambda = m + dfrac{1}{2}$ for $m in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
Thoughts -
Since $f(x)$ is continuous (multiplication of continuous functions) in $[-pi,pi]$ we could use Dirichlet's theorem and just say that $S$ should converge to $f(x)$, it just seems a little bit naive. What am i missing ?
Thanks in advance!
fourier-series harmonic-analysis
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
Let $f(x)=xcos(alpha x)$ at $[-pi,pi]$ for $lambda in mathbb{R}$. Let $S$ be $f(x)$'s corresponding Fourier Series.
Prove or disprove: 1) if $lambda in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
2) If $lambda = m + dfrac{1}{2}$ for $m in mathbb{Z}$ then $S$ Pointwise-converges to $f(x)$.
Thoughts -
Since $f(x)$ is continuous (multiplication of continuous functions) in $[-pi,pi]$ we could use Dirichlet's theorem and just say that $S$ should converge to $f(x)$, it just seems a little bit naive. What am i missing ?
Thanks in advance!
fourier-series harmonic-analysis
fourier-series harmonic-analysis
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
asked Nov 21 '18 at 22:18
Yariv LevyYariv Levy
7572416
7572416
You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47
add a comment |
You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47
You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47
You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47
add a comment |
1 Answer
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For any $lambda in mathbb R$ the function $f$ is continuous and it is of bounded variation. Hence the Fourier series converges to $f$ at every point. It is not true that the Fourier series of any continuous function converges pointwise. Dirichlet's Theorem only gives convergence almost everywhere, not everywhere.
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
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1 Answer
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For any $lambda in mathbb R$ the function $f$ is continuous and it is of bounded variation. Hence the Fourier series converges to $f$ at every point. It is not true that the Fourier series of any continuous function converges pointwise. Dirichlet's Theorem only gives convergence almost everywhere, not everywhere.
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
For any $lambda in mathbb R$ the function $f$ is continuous and it is of bounded variation. Hence the Fourier series converges to $f$ at every point. It is not true that the Fourier series of any continuous function converges pointwise. Dirichlet's Theorem only gives convergence almost everywhere, not everywhere.
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
For any $lambda in mathbb R$ the function $f$ is continuous and it is of bounded variation. Hence the Fourier series converges to $f$ at every point. It is not true that the Fourier series of any continuous function converges pointwise. Dirichlet's Theorem only gives convergence almost everywhere, not everywhere.
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
For any $lambda in mathbb R$ the function $f$ is continuous and it is of bounded variation. Hence the Fourier series converges to $f$ at every point. It is not true that the Fourier series of any continuous function converges pointwise. Dirichlet's Theorem only gives convergence almost everywhere, not everywhere.
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 7:28
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
add a comment |
add a comment |
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You are using a very deep theorem for an elementary result. Besides, pointwise here means at every point, not just almost everywhere. Dirichlet's Theorem only gives almost everywhere convergence.
– Kavi Rama Murthy
Nov 21 '18 at 23:47