Trigonometric System Completeness
This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $C(-pi,pi)$ instead of $C(0,2pi)$, but this makes no difference of course.
Suppose we had trigonometric polynomials $Q_k geq 0$ such that
- $$frac{1}{2pi}int_{-pi}^{pi} ! Q_k(t) , dt = 1$$
$Q_k to 0$ uniformly on $[-pi, -delta] cup [delta, pi]$ for every $delta >0$
(i.e. the $Q_k$ are a summation kernel). Associate to each continuous $f$ the function
$$P_k(t) = f ast Q_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(t-s)Q_k(s) , ds.$$
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial. Now let $epsilon > 0$. Since $f$ is uniformly continuous, there exists $delta > 0$ such that $|f(t) - f(s)| < epsilon$ whenever $|t-s| < delta$. Since each $Q_k$ has average equal to one, we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$
In the first term, the uniform continuity of $f$ shows that the integrand is smaller than $epsilon Q_k(s)$ and the assumption on the average of $Q_k$ makes this term small. For the second the integrand converges uniformly to $0$ as $k to infty$ (since $f$ is bounded). Since the estimates are independent of $t$ we have shown that
$$|P_k - f|_infty < epsilon$$
for $k$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $Q_k$ with the desired properties. To do this we let
$$Q_k(t) = c_k left(frac{1 + cos t}{2}right)^k$$
where the $c_k$ is chosen in such a way as to satisfy the assumption on the averages. Since the $Q_k$ are clearly positive, we only need to show that $Q_k to 0$ uniformly away from $0$. But $Q_k$ is even (on $(-pi,pi)$) and decreasing on $[0,pi]$. Thus for any $delta > 0$ we have
$$|Q_k(t)| le Q_k(delta) le frac{pi(k+1)}{2}left(frac{1 + cos delta}{2} right)^k to 0$$
independently of $t$ as $k to infty$ since
$$frac{1 + cos delta}{2} < 1.$$
My questions are as follows:
Why is it that $f$ is uniformly continuous?
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial.
we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$.
I think there are many details omitted in the proof. Thanks for the help in advance.
real-analysis fourier-analysis
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
add a comment |
This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $C(-pi,pi)$ instead of $C(0,2pi)$, but this makes no difference of course.
Suppose we had trigonometric polynomials $Q_k geq 0$ such that
- $$frac{1}{2pi}int_{-pi}^{pi} ! Q_k(t) , dt = 1$$
$Q_k to 0$ uniformly on $[-pi, -delta] cup [delta, pi]$ for every $delta >0$
(i.e. the $Q_k$ are a summation kernel). Associate to each continuous $f$ the function
$$P_k(t) = f ast Q_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(t-s)Q_k(s) , ds.$$
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial. Now let $epsilon > 0$. Since $f$ is uniformly continuous, there exists $delta > 0$ such that $|f(t) - f(s)| < epsilon$ whenever $|t-s| < delta$. Since each $Q_k$ has average equal to one, we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$
In the first term, the uniform continuity of $f$ shows that the integrand is smaller than $epsilon Q_k(s)$ and the assumption on the average of $Q_k$ makes this term small. For the second the integrand converges uniformly to $0$ as $k to infty$ (since $f$ is bounded). Since the estimates are independent of $t$ we have shown that
$$|P_k - f|_infty < epsilon$$
for $k$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $Q_k$ with the desired properties. To do this we let
$$Q_k(t) = c_k left(frac{1 + cos t}{2}right)^k$$
where the $c_k$ is chosen in such a way as to satisfy the assumption on the averages. Since the $Q_k$ are clearly positive, we only need to show that $Q_k to 0$ uniformly away from $0$. But $Q_k$ is even (on $(-pi,pi)$) and decreasing on $[0,pi]$. Thus for any $delta > 0$ we have
$$|Q_k(t)| le Q_k(delta) le frac{pi(k+1)}{2}left(frac{1 + cos delta}{2} right)^k to 0$$
independently of $t$ as $k to infty$ since
$$frac{1 + cos delta}{2} < 1.$$
My questions are as follows:
Why is it that $f$ is uniformly continuous?
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial.
we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$.
I think there are many details omitted in the proof. Thanks for the help in advance.
real-analysis fourier-analysis
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12
add a comment |
This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $C(-pi,pi)$ instead of $C(0,2pi)$, but this makes no difference of course.
Suppose we had trigonometric polynomials $Q_k geq 0$ such that
- $$frac{1}{2pi}int_{-pi}^{pi} ! Q_k(t) , dt = 1$$
$Q_k to 0$ uniformly on $[-pi, -delta] cup [delta, pi]$ for every $delta >0$
(i.e. the $Q_k$ are a summation kernel). Associate to each continuous $f$ the function
$$P_k(t) = f ast Q_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(t-s)Q_k(s) , ds.$$
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial. Now let $epsilon > 0$. Since $f$ is uniformly continuous, there exists $delta > 0$ such that $|f(t) - f(s)| < epsilon$ whenever $|t-s| < delta$. Since each $Q_k$ has average equal to one, we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$
In the first term, the uniform continuity of $f$ shows that the integrand is smaller than $epsilon Q_k(s)$ and the assumption on the average of $Q_k$ makes this term small. For the second the integrand converges uniformly to $0$ as $k to infty$ (since $f$ is bounded). Since the estimates are independent of $t$ we have shown that
$$|P_k - f|_infty < epsilon$$
for $k$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $Q_k$ with the desired properties. To do this we let
$$Q_k(t) = c_k left(frac{1 + cos t}{2}right)^k$$
where the $c_k$ is chosen in such a way as to satisfy the assumption on the averages. Since the $Q_k$ are clearly positive, we only need to show that $Q_k to 0$ uniformly away from $0$. But $Q_k$ is even (on $(-pi,pi)$) and decreasing on $[0,pi]$. Thus for any $delta > 0$ we have
$$|Q_k(t)| le Q_k(delta) le frac{pi(k+1)}{2}left(frac{1 + cos delta}{2} right)^k to 0$$
independently of $t$ as $k to infty$ since
$$frac{1 + cos delta}{2} < 1.$$
My questions are as follows:
Why is it that $f$ is uniformly continuous?
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial.
we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$.
I think there are many details omitted in the proof. Thanks for the help in advance.
real-analysis fourier-analysis
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
This argument is from Rudin's Real and Complex Analysis (section 4.24 Completeness of the Trigonometric System). It considers $C(-pi,pi)$ instead of $C(0,2pi)$, but this makes no difference of course.
Suppose we had trigonometric polynomials $Q_k geq 0$ such that
- $$frac{1}{2pi}int_{-pi}^{pi} ! Q_k(t) , dt = 1$$
$Q_k to 0$ uniformly on $[-pi, -delta] cup [delta, pi]$ for every $delta >0$
(i.e. the $Q_k$ are a summation kernel). Associate to each continuous $f$ the function
$$P_k(t) = f ast Q_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(t-s)Q_k(s) , ds.$$
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial. Now let $epsilon > 0$. Since $f$ is uniformly continuous, there exists $delta > 0$ such that $|f(t) - f(s)| < epsilon$ whenever $|t-s| < delta$. Since each $Q_k$ has average equal to one, we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$
In the first term, the uniform continuity of $f$ shows that the integrand is smaller than $epsilon Q_k(s)$ and the assumption on the average of $Q_k$ makes this term small. For the second the integrand converges uniformly to $0$ as $k to infty$ (since $f$ is bounded). Since the estimates are independent of $t$ we have shown that
$$|P_k - f|_infty < epsilon$$
for $k$ large enough and thus the trigonometric polynomials are dense. We need to now construct the $Q_k$ with the desired properties. To do this we let
$$Q_k(t) = c_k left(frac{1 + cos t}{2}right)^k$$
where the $c_k$ is chosen in such a way as to satisfy the assumption on the averages. Since the $Q_k$ are clearly positive, we only need to show that $Q_k to 0$ uniformly away from $0$. But $Q_k$ is even (on $(-pi,pi)$) and decreasing on $[0,pi]$. Thus for any $delta > 0$ we have
$$|Q_k(t)| le Q_k(delta) le frac{pi(k+1)}{2}left(frac{1 + cos delta}{2} right)^k to 0$$
independently of $t$ as $k to infty$ since
$$frac{1 + cos delta}{2} < 1.$$
My questions are as follows:
Why is it that $f$ is uniformly continuous?
Substitution shows that also
$$P_k(t) = frac{1}{2pi}int_{-pi}^pi ! f(s)Q_k(t-s) , ds$$
so that each $P_k$ is a trigonometric polynomial.
we see that
$$P_k(t) - f(t) = frac{1}{2pi}int_{-pi}^pi ! (f(t-s) - f(s)) Q_k(s) , ds$$ and the positivity of $Q_k$ implies
$$|P_k(t) - f(t)| le frac{1}{2pi}int_{-pi}^pi ! |f(t-s) - f(s)|Q_k(s) , ds$$
$$= frac{1}{2pi}int_{|s| le delta} ! |f(t-s) - f(s)|Q_k(s) , ds + frac{1}{2pi}int_{|s|geq delta} ! |f(t-s) - f(s)|Q_k(s) , ds$$.
I think there are many details omitted in the proof. Thanks for the help in advance.
real-analysis fourier-analysis
real-analysis fourier-analysis
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
asked Dec 5 '18 at 16:32
Heisenberg Heisenberg
75
75
$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12
add a comment |
$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12
$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12
add a comment |
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$f$ is continuous by assumption and defined on the compact set $[-pi,pi]$. Continuity on a compact set implies uniform continuity.
– Paul Sinclair
Dec 6 '18 at 0:16
How about the other?
– Heisenberg
Dec 6 '18 at 3:52
The higher the math is, the fewer details you see in proofs. The proof is a map showing the way, but you have walk the path yourself. Without more information about what you don't understand, I can't say anything more.
– Paul Sinclair
Dec 6 '18 at 4:12