Mixing
If $fin BV(mathbb{T})cap C(mathbb{T})$ does the Fourier series of $f$ converge uniformly to $f$?
$begingroup$
Denote with $BV(mathbb{T})$ the set of the functions of bounded variation defined on the 1-torus $mathbb{T}$.
If $fin BV(mathbb{T})$, define
$$f^°:mathbb{T}tomathbb{C}, tmapsto frac{lim_{sto t^+}f(s)+lim_{sto t^-}f(s)}{2}.$$
If $fin L^1(mathbb{T})$, define:
$$hat{f}:mathbb{Z}tomathbb{C}, nmapstoint_{-pi}^{pi}f(t)e^{-int}frac{operatorname{d}t}{2pi}$$
Then (see Duoandikoetxea - Fourier Analysis, Theorem 1.2):
$$forall fin BV(mathbb{T}), forall tinmathbb{T}, sum_{n=-N}^Nhat{f}(n)e^{int}to f^°(t), Ntoinfty.$$
Obviosly, we can't expect much more then pointwise convergence if $fin BV(mathbb{T})$ because uniform convergence would imply that $fin C(mathbb{T})$ while in general this is not the case. However, if $fin BV(mathbb{T})cap C(mathbb{T})$, then $f^°=f$, and so the question makes sense, i.e.
Is it true that if $fin BV(mathbb{T})cap C(mathbb{T})$ then $sup_{tinmathbb{T}}|sum_{n=-N}^Nhat{f}(n)e^{int}- f(t)|to 0, Ntoinfty?$
fourier-series
asked Jan 22 at 10:26
BobBob
1,7311725
$endgroup$
add a comment |
$begingroup$
Denote with $BV(mathbb{T})$ the set of the functions of bounded variation defined on the 1-torus $mathbb{T}$.
If $fin BV(mathbb{T})$, define
$$f^°:mathbb{T}tomathbb{C}, tmapsto frac{lim_{sto t^+}f(s)+lim_{sto t^-}f(s)}{2}.$$
If $fin L^1(mathbb{T})$, define:
$$hat{f}:mathbb{Z}tomathbb{C}, nmapstoint_{-pi}^{pi}f(t)e^{-int}frac{operatorname{d}t}{2pi}$$
Then (see Duoandikoetxea - Fourier Analysis, Theorem 1.2):
$$forall fin BV(mathbb{T}), forall tinmathbb{T}, sum_{n=-N}^Nhat{f}(n)e^{int}to f^°(t), Ntoinfty.$$
Obviosly, we can't expect much more then pointwise convergence if $fin BV(mathbb{T})$ because uniform convergence would imply that $fin C(mathbb{T})$ while in general this is not the case. However, if $fin BV(mathbb{T})cap C(mathbb{T})$, then $f^°=f$, and so the question makes sense, i.e.
Is it true that if $fin BV(mathbb{T})cap C(mathbb{T})$ then $sup_{tinmathbb{T}}|sum_{n=-N}^Nhat{f}(n)e^{int}- f(t)|to 0, Ntoinfty?$
fourier-series
asked Jan 22 at 10:26
BobBob
1,7311725
$endgroup$
2
$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28
add a comment |
$begingroup$
Denote with $BV(mathbb{T})$ the set of the functions of bounded variation defined on the 1-torus $mathbb{T}$.
If $fin BV(mathbb{T})$, define
$$f^°:mathbb{T}tomathbb{C}, tmapsto frac{lim_{sto t^+}f(s)+lim_{sto t^-}f(s)}{2}.$$
If $fin L^1(mathbb{T})$, define:
$$hat{f}:mathbb{Z}tomathbb{C}, nmapstoint_{-pi}^{pi}f(t)e^{-int}frac{operatorname{d}t}{2pi}$$
Then (see Duoandikoetxea - Fourier Analysis, Theorem 1.2):
$$forall fin BV(mathbb{T}), forall tinmathbb{T}, sum_{n=-N}^Nhat{f}(n)e^{int}to f^°(t), Ntoinfty.$$
Obviosly, we can't expect much more then pointwise convergence if $fin BV(mathbb{T})$ because uniform convergence would imply that $fin C(mathbb{T})$ while in general this is not the case. However, if $fin BV(mathbb{T})cap C(mathbb{T})$, then $f^°=f$, and so the question makes sense, i.e.
Is it true that if $fin BV(mathbb{T})cap C(mathbb{T})$ then $sup_{tinmathbb{T}}|sum_{n=-N}^Nhat{f}(n)e^{int}- f(t)|to 0, Ntoinfty?$
fourier-series
asked Jan 22 at 10:26
BobBob
1,7311725
$endgroup$
Denote with $BV(mathbb{T})$ the set of the functions of bounded variation defined on the 1-torus $mathbb{T}$.
If $fin BV(mathbb{T})$, define
$$f^°:mathbb{T}tomathbb{C}, tmapsto frac{lim_{sto t^+}f(s)+lim_{sto t^-}f(s)}{2}.$$
If $fin L^1(mathbb{T})$, define:
$$hat{f}:mathbb{Z}tomathbb{C}, nmapstoint_{-pi}^{pi}f(t)e^{-int}frac{operatorname{d}t}{2pi}$$
Then (see Duoandikoetxea - Fourier Analysis, Theorem 1.2):
$$forall fin BV(mathbb{T}), forall tinmathbb{T}, sum_{n=-N}^Nhat{f}(n)e^{int}to f^°(t), Ntoinfty.$$
Obviosly, we can't expect much more then pointwise convergence if $fin BV(mathbb{T})$ because uniform convergence would imply that $fin C(mathbb{T})$ while in general this is not the case. However, if $fin BV(mathbb{T})cap C(mathbb{T})$, then $f^°=f$, and so the question makes sense, i.e.
Is it true that if $fin BV(mathbb{T})cap C(mathbb{T})$ then $sup_{tinmathbb{T}}|sum_{n=-N}^Nhat{f}(n)e^{int}- f(t)|to 0, Ntoinfty?$
fourier-series
fourier-series
asked Jan 22 at 10:26
BobBob
1,7311725
asked Jan 22 at 10:26
BobBob
1,7311725
edited Jan 22 at 10:29
Bob
asked Jan 22 at 10:26
BobBob
1,7311725
asked Jan 22 at 10:26
BobBob
1,7311725
asked Jan 22 at 10:26
BobBob
1,7311725
1,7311725
2
$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28
add a comment |
2
$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28
2
2
$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28
$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:mathbb{R}tomathbb{R}$ is a $2pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $mu_f$, i.e. $mu_f$ is the only signed measure such that
$$mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:mathbb{R}to mathbb{R}$ as a variation of $f$, i.e.
$$|mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $varphiin C_c^1(mathbb{R})$ then it holds the follow integration by parts formula:
$$int_mathbb{R}f(t)varphi'(t)operatorname{d}t=-int_mathbb{R}varphi(t)operatorname{d}mu_f(t)$$
that leads to:
$$forallvarphiin C^1(mathbb{R}), forall ainmathbb{R}, forall b>a, int_a^bf(t)varphi'(t)operatorname{d}t=f(b^+)varphi(b)-f(a^-)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$mu _f([a,b])=f(b)-f(a) \ |mu _f|([a,b])=V_f(b)-V_f(a)\ int_a^bf(t)varphi'(t)operatorname{d}t=f(b)varphi(b)-f(a)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $mu_{f_x}(A)=mu_{f}(x+A)$, $|mu_{f_x}|(A)=|mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$varphi_N(s):=int_0^s frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}operatorname{d}t$$
In the answer to this question it is proved that:
$$exists C>0, forall sin[-pi,pi], forall Ninmathbb{N}, |varphi_N(s)|le C.$$
Second: a reformulation of what we want to prove
We have that:
$$sum_{n=-N}^N hat{f}(n)e^{inx}-f(x)=int_{-pi}^pi(f(x+t)-f(x))frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi} = int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi},$$
so we want to prove that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto +infty.$$
Now:
If $deltain(0,pi)$ we have that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|\ le sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right| + sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|.$$
So it enough to prove that for all $varepsilon>0$ there exists $deltain(0,pi)$ such that:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
and:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to0, Nto+infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|=left|int_{-delta}^delta f_x(t) varphi_N'(t)frac{operatorname{d}t}{2pi}right|\ = frac{1}{2pi}left|-int_{[-delta,delta]} varphi_N(t)operatorname{d}mu_{f_x}(t)+f_x(delta)varphi_N(delta)-f_x(-delta)varphi_N(-delta)right|le frac{C}{pi} |mu_{f_x}|([-delta,delta]) \ = frac{C}{pi} |mu_{f}|([x-delta,x+delta]) = frac{C}{pi} (V_f(x+delta)-V_f(x+delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2pi,2pi]$. So, if $varepsilon>0$ and $deltain(0,frac{pi}{2})$ is such that for $|x-y|le 2delta$ we have that $|V_f(x)-V_f(y)|<varepsilon$, then we have that:
$$forall xin [-pi,pi], (V_f(x+delta)-V_f(x+delta)) <varepsilon$$
and so:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
Fourth: second integral estimate
Since:
$$sinleft((N+frac{1}{2})tright)= sin (Nt) cos(frac{t}{2})+cos (Nt) sin(frac{t}{2}),$$ we have that:
$$frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}=frac{sin(Nt)}{tan(frac{t}{2})}+cos(Nt).$$
Now:
$$int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}t = -int_{-pi}^{pi} f_x(t-frac{pi}{N}) cos(Nt) operatorname{d}t,$$
so:
$$left|int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}tright| = left|frac{1}{2} int_{-pi}^{pi} (f_x(t)-f_x(t-frac{pi}{N})) cos(Nt) operatorname{d}tright| = left| frac{1}{2} int_{-pi}^{pi} (f(t)-f(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|le frac{1}{2} int_{-pi}^{pi} left|(f(t)-f(t-frac{pi}{N}))right| operatorname{d}tle omega_{f,1}(frac{pi}{N}) $$
where $$omega_{g,1}(alpha)=sup_{hin[-alpha,alpha]}int_{-pi}^{pi} |g(t+h)-g(t)|operatorname{d}t$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|le sup_{xin[-pi,pi]}left|int_{[-pi,pi]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|
le frac{1}{2pi} omega_{f,1}(frac{pi}{N})$$
So, since $fin L^1([-pi,pi])$, we have that
$$omega_{f,1}(frac{pi}{N})to 0, Nto+infty$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
So, it remains to prove that:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin(Nt)}{tan(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
Now, let $psi$ be a continuous $2pi$-function that coincides with $tmapsto frac{1}{tan(frac{t}{2})}$ on $[-pi,pi]backslash[-delta,delta]$. Then, what we want to prove is equivalent to:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
With the same technique as before, we have that:
$$left|int_{[-pi,pi]} f_x(t)psi(t)sin(Nt){operatorname{d}t}right|le left|frac{1}{2} int_{-pi}^{pi} (f_x(t)psi(t)-f_x(t-frac{pi}{N}))psi(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|\ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)+f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)right|operatorname{d}t+ frac{1}{2} int_{-pi}^{pi}left|f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} int_{-pi}^{pi}left| f_x(t)-f_x(t-frac{pi}{N})right|operatorname{d}t+ frac{|f|_infty}{2} int_{-pi}^{pi}left|psi(t)-psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})$$
and since $f,psiin L^1([-pi,pi])$
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|lefrac{1}{2pi}left(frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})right)to 0, Nto+infty.$$
answered Jan 23 at 22:13
BobBob
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$begingroup$
Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:mathbb{R}tomathbb{R}$ is a $2pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $mu_f$, i.e. $mu_f$ is the only signed measure such that
$$mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:mathbb{R}to mathbb{R}$ as a variation of $f$, i.e.
$$|mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $varphiin C_c^1(mathbb{R})$ then it holds the follow integration by parts formula:
$$int_mathbb{R}f(t)varphi'(t)operatorname{d}t=-int_mathbb{R}varphi(t)operatorname{d}mu_f(t)$$
that leads to:
$$forallvarphiin C^1(mathbb{R}), forall ainmathbb{R}, forall b>a, int_a^bf(t)varphi'(t)operatorname{d}t=f(b^+)varphi(b)-f(a^-)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$mu _f([a,b])=f(b)-f(a) \ |mu _f|([a,b])=V_f(b)-V_f(a)\ int_a^bf(t)varphi'(t)operatorname{d}t=f(b)varphi(b)-f(a)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $mu_{f_x}(A)=mu_{f}(x+A)$, $|mu_{f_x}|(A)=|mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$varphi_N(s):=int_0^s frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}operatorname{d}t$$
In the answer to this question it is proved that:
$$exists C>0, forall sin[-pi,pi], forall Ninmathbb{N}, |varphi_N(s)|le C.$$
Second: a reformulation of what we want to prove
We have that:
$$sum_{n=-N}^N hat{f}(n)e^{inx}-f(x)=int_{-pi}^pi(f(x+t)-f(x))frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi} = int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi},$$
so we want to prove that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto +infty.$$
Now:
If $deltain(0,pi)$ we have that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|\ le sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right| + sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|.$$
So it enough to prove that for all $varepsilon>0$ there exists $deltain(0,pi)$ such that:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
and:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to0, Nto+infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|=left|int_{-delta}^delta f_x(t) varphi_N'(t)frac{operatorname{d}t}{2pi}right|\ = frac{1}{2pi}left|-int_{[-delta,delta]} varphi_N(t)operatorname{d}mu_{f_x}(t)+f_x(delta)varphi_N(delta)-f_x(-delta)varphi_N(-delta)right|le frac{C}{pi} |mu_{f_x}|([-delta,delta]) \ = frac{C}{pi} |mu_{f}|([x-delta,x+delta]) = frac{C}{pi} (V_f(x+delta)-V_f(x+delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2pi,2pi]$. So, if $varepsilon>0$ and $deltain(0,frac{pi}{2})$ is such that for $|x-y|le 2delta$ we have that $|V_f(x)-V_f(y)|<varepsilon$, then we have that:
$$forall xin [-pi,pi], (V_f(x+delta)-V_f(x+delta)) <varepsilon$$
and so:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
Fourth: second integral estimate
Since:
$$sinleft((N+frac{1}{2})tright)= sin (Nt) cos(frac{t}{2})+cos (Nt) sin(frac{t}{2}),$$ we have that:
$$frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}=frac{sin(Nt)}{tan(frac{t}{2})}+cos(Nt).$$
Now:
$$int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}t = -int_{-pi}^{pi} f_x(t-frac{pi}{N}) cos(Nt) operatorname{d}t,$$
so:
$$left|int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}tright| = left|frac{1}{2} int_{-pi}^{pi} (f_x(t)-f_x(t-frac{pi}{N})) cos(Nt) operatorname{d}tright| = left| frac{1}{2} int_{-pi}^{pi} (f(t)-f(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|le frac{1}{2} int_{-pi}^{pi} left|(f(t)-f(t-frac{pi}{N}))right| operatorname{d}tle omega_{f,1}(frac{pi}{N}) $$
where $$omega_{g,1}(alpha)=sup_{hin[-alpha,alpha]}int_{-pi}^{pi} |g(t+h)-g(t)|operatorname{d}t$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|le sup_{xin[-pi,pi]}left|int_{[-pi,pi]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|
le frac{1}{2pi} omega_{f,1}(frac{pi}{N})$$
So, since $fin L^1([-pi,pi])$, we have that
$$omega_{f,1}(frac{pi}{N})to 0, Nto+infty$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
So, it remains to prove that:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin(Nt)}{tan(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
Now, let $psi$ be a continuous $2pi$-function that coincides with $tmapsto frac{1}{tan(frac{t}{2})}$ on $[-pi,pi]backslash[-delta,delta]$. Then, what we want to prove is equivalent to:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
With the same technique as before, we have that:
$$left|int_{[-pi,pi]} f_x(t)psi(t)sin(Nt){operatorname{d}t}right|le left|frac{1}{2} int_{-pi}^{pi} (f_x(t)psi(t)-f_x(t-frac{pi}{N}))psi(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|\ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)+f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)right|operatorname{d}t+ frac{1}{2} int_{-pi}^{pi}left|f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} int_{-pi}^{pi}left| f_x(t)-f_x(t-frac{pi}{N})right|operatorname{d}t+ frac{|f|_infty}{2} int_{-pi}^{pi}left|psi(t)-psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})$$
and since $f,psiin L^1([-pi,pi])$
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|lefrac{1}{2pi}left(frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})right)to 0, Nto+infty.$$
answered Jan 23 at 22:13
BobBob
1,7311725
$endgroup$
add a comment |
$begingroup$
Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:mathbb{R}tomathbb{R}$ is a $2pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $mu_f$, i.e. $mu_f$ is the only signed measure such that
$$mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:mathbb{R}to mathbb{R}$ as a variation of $f$, i.e.
$$|mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $varphiin C_c^1(mathbb{R})$ then it holds the follow integration by parts formula:
$$int_mathbb{R}f(t)varphi'(t)operatorname{d}t=-int_mathbb{R}varphi(t)operatorname{d}mu_f(t)$$
that leads to:
$$forallvarphiin C^1(mathbb{R}), forall ainmathbb{R}, forall b>a, int_a^bf(t)varphi'(t)operatorname{d}t=f(b^+)varphi(b)-f(a^-)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$mu _f([a,b])=f(b)-f(a) \ |mu _f|([a,b])=V_f(b)-V_f(a)\ int_a^bf(t)varphi'(t)operatorname{d}t=f(b)varphi(b)-f(a)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $mu_{f_x}(A)=mu_{f}(x+A)$, $|mu_{f_x}|(A)=|mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$varphi_N(s):=int_0^s frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}operatorname{d}t$$
In the answer to this question it is proved that:
$$exists C>0, forall sin[-pi,pi], forall Ninmathbb{N}, |varphi_N(s)|le C.$$
Second: a reformulation of what we want to prove
We have that:
$$sum_{n=-N}^N hat{f}(n)e^{inx}-f(x)=int_{-pi}^pi(f(x+t)-f(x))frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi} = int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi},$$
so we want to prove that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto +infty.$$
Now:
If $deltain(0,pi)$ we have that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|\ le sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right| + sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|.$$
So it enough to prove that for all $varepsilon>0$ there exists $deltain(0,pi)$ such that:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
and:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to0, Nto+infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|=left|int_{-delta}^delta f_x(t) varphi_N'(t)frac{operatorname{d}t}{2pi}right|\ = frac{1}{2pi}left|-int_{[-delta,delta]} varphi_N(t)operatorname{d}mu_{f_x}(t)+f_x(delta)varphi_N(delta)-f_x(-delta)varphi_N(-delta)right|le frac{C}{pi} |mu_{f_x}|([-delta,delta]) \ = frac{C}{pi} |mu_{f}|([x-delta,x+delta]) = frac{C}{pi} (V_f(x+delta)-V_f(x+delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2pi,2pi]$. So, if $varepsilon>0$ and $deltain(0,frac{pi}{2})$ is such that for $|x-y|le 2delta$ we have that $|V_f(x)-V_f(y)|<varepsilon$, then we have that:
$$forall xin [-pi,pi], (V_f(x+delta)-V_f(x+delta)) <varepsilon$$
and so:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
Fourth: second integral estimate
Since:
$$sinleft((N+frac{1}{2})tright)= sin (Nt) cos(frac{t}{2})+cos (Nt) sin(frac{t}{2}),$$ we have that:
$$frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}=frac{sin(Nt)}{tan(frac{t}{2})}+cos(Nt).$$
Now:
$$int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}t = -int_{-pi}^{pi} f_x(t-frac{pi}{N}) cos(Nt) operatorname{d}t,$$
so:
$$left|int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}tright| = left|frac{1}{2} int_{-pi}^{pi} (f_x(t)-f_x(t-frac{pi}{N})) cos(Nt) operatorname{d}tright| = left| frac{1}{2} int_{-pi}^{pi} (f(t)-f(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|le frac{1}{2} int_{-pi}^{pi} left|(f(t)-f(t-frac{pi}{N}))right| operatorname{d}tle omega_{f,1}(frac{pi}{N}) $$
where $$omega_{g,1}(alpha)=sup_{hin[-alpha,alpha]}int_{-pi}^{pi} |g(t+h)-g(t)|operatorname{d}t$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|le sup_{xin[-pi,pi]}left|int_{[-pi,pi]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|
le frac{1}{2pi} omega_{f,1}(frac{pi}{N})$$
So, since $fin L^1([-pi,pi])$, we have that
$$omega_{f,1}(frac{pi}{N})to 0, Nto+infty$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
So, it remains to prove that:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin(Nt)}{tan(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
Now, let $psi$ be a continuous $2pi$-function that coincides with $tmapsto frac{1}{tan(frac{t}{2})}$ on $[-pi,pi]backslash[-delta,delta]$. Then, what we want to prove is equivalent to:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
With the same technique as before, we have that:
$$left|int_{[-pi,pi]} f_x(t)psi(t)sin(Nt){operatorname{d}t}right|le left|frac{1}{2} int_{-pi}^{pi} (f_x(t)psi(t)-f_x(t-frac{pi}{N}))psi(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|\ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)+f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)right|operatorname{d}t+ frac{1}{2} int_{-pi}^{pi}left|f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} int_{-pi}^{pi}left| f_x(t)-f_x(t-frac{pi}{N})right|operatorname{d}t+ frac{|f|_infty}{2} int_{-pi}^{pi}left|psi(t)-psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})$$
and since $f,psiin L^1([-pi,pi])$
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|lefrac{1}{2pi}left(frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})right)to 0, Nto+infty.$$
answered Jan 23 at 22:13
BobBob
1,7311725
$endgroup$
add a comment |
$begingroup$
Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:mathbb{R}tomathbb{R}$ is a $2pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $mu_f$, i.e. $mu_f$ is the only signed measure such that
$$mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:mathbb{R}to mathbb{R}$ as a variation of $f$, i.e.
$$|mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $varphiin C_c^1(mathbb{R})$ then it holds the follow integration by parts formula:
$$int_mathbb{R}f(t)varphi'(t)operatorname{d}t=-int_mathbb{R}varphi(t)operatorname{d}mu_f(t)$$
that leads to:
$$forallvarphiin C^1(mathbb{R}), forall ainmathbb{R}, forall b>a, int_a^bf(t)varphi'(t)operatorname{d}t=f(b^+)varphi(b)-f(a^-)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$mu _f([a,b])=f(b)-f(a) \ |mu _f|([a,b])=V_f(b)-V_f(a)\ int_a^bf(t)varphi'(t)operatorname{d}t=f(b)varphi(b)-f(a)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $mu_{f_x}(A)=mu_{f}(x+A)$, $|mu_{f_x}|(A)=|mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$varphi_N(s):=int_0^s frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}operatorname{d}t$$
In the answer to this question it is proved that:
$$exists C>0, forall sin[-pi,pi], forall Ninmathbb{N}, |varphi_N(s)|le C.$$
Second: a reformulation of what we want to prove
We have that:
$$sum_{n=-N}^N hat{f}(n)e^{inx}-f(x)=int_{-pi}^pi(f(x+t)-f(x))frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi} = int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi},$$
so we want to prove that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto +infty.$$
Now:
If $deltain(0,pi)$ we have that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|\ le sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right| + sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|.$$
So it enough to prove that for all $varepsilon>0$ there exists $deltain(0,pi)$ such that:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
and:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to0, Nto+infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|=left|int_{-delta}^delta f_x(t) varphi_N'(t)frac{operatorname{d}t}{2pi}right|\ = frac{1}{2pi}left|-int_{[-delta,delta]} varphi_N(t)operatorname{d}mu_{f_x}(t)+f_x(delta)varphi_N(delta)-f_x(-delta)varphi_N(-delta)right|le frac{C}{pi} |mu_{f_x}|([-delta,delta]) \ = frac{C}{pi} |mu_{f}|([x-delta,x+delta]) = frac{C}{pi} (V_f(x+delta)-V_f(x+delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2pi,2pi]$. So, if $varepsilon>0$ and $deltain(0,frac{pi}{2})$ is such that for $|x-y|le 2delta$ we have that $|V_f(x)-V_f(y)|<varepsilon$, then we have that:
$$forall xin [-pi,pi], (V_f(x+delta)-V_f(x+delta)) <varepsilon$$
and so:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
Fourth: second integral estimate
Since:
$$sinleft((N+frac{1}{2})tright)= sin (Nt) cos(frac{t}{2})+cos (Nt) sin(frac{t}{2}),$$ we have that:
$$frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}=frac{sin(Nt)}{tan(frac{t}{2})}+cos(Nt).$$
Now:
$$int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}t = -int_{-pi}^{pi} f_x(t-frac{pi}{N}) cos(Nt) operatorname{d}t,$$
so:
$$left|int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}tright| = left|frac{1}{2} int_{-pi}^{pi} (f_x(t)-f_x(t-frac{pi}{N})) cos(Nt) operatorname{d}tright| = left| frac{1}{2} int_{-pi}^{pi} (f(t)-f(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|le frac{1}{2} int_{-pi}^{pi} left|(f(t)-f(t-frac{pi}{N}))right| operatorname{d}tle omega_{f,1}(frac{pi}{N}) $$
where $$omega_{g,1}(alpha)=sup_{hin[-alpha,alpha]}int_{-pi}^{pi} |g(t+h)-g(t)|operatorname{d}t$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|le sup_{xin[-pi,pi]}left|int_{[-pi,pi]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|
le frac{1}{2pi} omega_{f,1}(frac{pi}{N})$$
So, since $fin L^1([-pi,pi])$, we have that
$$omega_{f,1}(frac{pi}{N})to 0, Nto+infty$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
So, it remains to prove that:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin(Nt)}{tan(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
Now, let $psi$ be a continuous $2pi$-function that coincides with $tmapsto frac{1}{tan(frac{t}{2})}$ on $[-pi,pi]backslash[-delta,delta]$. Then, what we want to prove is equivalent to:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
With the same technique as before, we have that:
$$left|int_{[-pi,pi]} f_x(t)psi(t)sin(Nt){operatorname{d}t}right|le left|frac{1}{2} int_{-pi}^{pi} (f_x(t)psi(t)-f_x(t-frac{pi}{N}))psi(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|\ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)+f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)right|operatorname{d}t+ frac{1}{2} int_{-pi}^{pi}left|f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} int_{-pi}^{pi}left| f_x(t)-f_x(t-frac{pi}{N})right|operatorname{d}t+ frac{|f|_infty}{2} int_{-pi}^{pi}left|psi(t)-psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})$$
and since $f,psiin L^1([-pi,pi])$
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|lefrac{1}{2pi}left(frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})right)to 0, Nto+infty.$$
answered Jan 23 at 22:13
BobBob
1,7311725
$endgroup$
Let's prove the theorem for real-valued functions.
First: notations and useful results
If $f:mathbb{R}tomathbb{R}$ is a $2pi$-periodic function continuous from the right of bounded variation over a period, denote the Lebesgue-Stieltjes signed measure associated to $f$ with $mu_f$, i.e. $mu_f$ is the only signed measure such that
$$mu _f((a,b])=f(b)-f(a).$$
Also, define $V_f:mathbb{R}to mathbb{R}$ as a variation of $f$, i.e.
$$|mu_f|((a,b])=V_f(b)-V_f(a).$$
Recall that if $varphiin C_c^1(mathbb{R})$ then it holds the follow integration by parts formula:
$$int_mathbb{R}f(t)varphi'(t)operatorname{d}t=-int_mathbb{R}varphi(t)operatorname{d}mu_f(t)$$
that leads to:
$$forallvarphiin C^1(mathbb{R}), forall ainmathbb{R}, forall b>a, int_a^bf(t)varphi'(t)operatorname{d}t=f(b^+)varphi(b)-f(a^-)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
In our case, $f$ is also continuous, so actually we also have that:
$$mu _f([a,b])=f(b)-f(a) \ |mu _f|([a,b])=V_f(b)-V_f(a)\ int_a^bf(t)varphi'(t)operatorname{d}t=f(b)varphi(b)-f(a)varphi(a)-int_{[a,b]}varphi(t)operatorname{d}mu_f(t).$$
Also, define $f_x(t):=f(x+t)-f(x)$ and notice that $mu_{f_x}(A)=mu_{f}(x+A)$, $|mu_{f_x}|(A)=|mu_{f}|(x+A)$ and $V_{f_x}(t)=V_f(x+t).$
Also, define $$varphi_N(s):=int_0^s frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}operatorname{d}t$$
In the answer to this question it is proved that:
$$exists C>0, forall sin[-pi,pi], forall Ninmathbb{N}, |varphi_N(s)|le C.$$
Second: a reformulation of what we want to prove
We have that:
$$sum_{n=-N}^N hat{f}(n)e^{inx}-f(x)=int_{-pi}^pi(f(x+t)-f(x))frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi} = int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi},$$
so we want to prove that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto +infty.$$
Now:
If $deltain(0,pi)$ we have that:
$$sup_{xin[-pi,pi]}left|int_{-pi}^pi f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|\ le sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right| + sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|.$$
So it enough to prove that for all $varepsilon>0$ there exists $deltain(0,pi)$ such that:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
and:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to0, Nto+infty.$$
Third: first integral estimate
Let's use the integration by parts formula:
$$left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|=left|int_{-delta}^delta f_x(t) varphi_N'(t)frac{operatorname{d}t}{2pi}right|\ = frac{1}{2pi}left|-int_{[-delta,delta]} varphi_N(t)operatorname{d}mu_{f_x}(t)+f_x(delta)varphi_N(delta)-f_x(-delta)varphi_N(-delta)right|le frac{C}{pi} |mu_{f_x}|([-delta,delta]) \ = frac{C}{pi} |mu_{f}|([x-delta,x+delta]) = frac{C}{pi} (V_f(x+delta)-V_f(x+delta)).$$
Now, being $f$ continuous, we have that $V_f$ is continuous and so uniformly continuous e.g. over the interval $[-2pi,2pi]$. So, if $varepsilon>0$ and $deltain(0,frac{pi}{2})$ is such that for $|x-y|le 2delta$ we have that $|V_f(x)-V_f(y)|<varepsilon$, then we have that:
$$forall xin [-pi,pi], (V_f(x+delta)-V_f(x+delta)) <varepsilon$$
and so:
$$forall Ninmathbb{N}, sup_{xin[-pi,pi]}left|int_{-delta}^delta f_x(t)frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}frac{operatorname{d}t}{2pi}right|le frac{C}{pi}varepsilon$$
Fourth: second integral estimate
Since:
$$sinleft((N+frac{1}{2})tright)= sin (Nt) cos(frac{t}{2})+cos (Nt) sin(frac{t}{2}),$$ we have that:
$$frac{sin((N+frac{1}{2})t)}{sin(frac{t}{2})}=frac{sin(Nt)}{tan(frac{t}{2})}+cos(Nt).$$
Now:
$$int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}t = -int_{-pi}^{pi} f_x(t-frac{pi}{N}) cos(Nt) operatorname{d}t,$$
so:
$$left|int_{-pi}^{pi} f_x(t) cos(Nt) operatorname{d}tright| = left|frac{1}{2} int_{-pi}^{pi} (f_x(t)-f_x(t-frac{pi}{N})) cos(Nt) operatorname{d}tright| = left| frac{1}{2} int_{-pi}^{pi} (f(t)-f(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|le frac{1}{2} int_{-pi}^{pi} left|(f(t)-f(t-frac{pi}{N}))right| operatorname{d}tle omega_{f,1}(frac{pi}{N}) $$
where $$omega_{g,1}(alpha)=sup_{hin[-alpha,alpha]}int_{-pi}^{pi} |g(t+h)-g(t)|operatorname{d}t$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|le sup_{xin[-pi,pi]}left|int_{[-pi,pi]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|
le frac{1}{2pi} omega_{f,1}(frac{pi}{N})$$
So, since $fin L^1([-pi,pi])$, we have that
$$omega_{f,1}(frac{pi}{N})to 0, Nto+infty$$
and then:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)cos(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
So, it remains to prove that:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)frac{sin(Nt)}{tan(frac{t}{2})}frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
Now, let $psi$ be a continuous $2pi$-function that coincides with $tmapsto frac{1}{tan(frac{t}{2})}$ on $[-pi,pi]backslash[-delta,delta]$. Then, what we want to prove is equivalent to:
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|to 0, Nto+infty.$$
With the same technique as before, we have that:
$$left|int_{[-pi,pi]} f_x(t)psi(t)sin(Nt){operatorname{d}t}right|le left|frac{1}{2} int_{-pi}^{pi} (f_x(t)psi(t)-f_x(t-frac{pi}{N}))psi(t-frac{pi}{N})) cos(Nt) operatorname{d}tright|\ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)+f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{1}{2} int_{-pi}^{pi}left| f_x(t)psi(t)-f_x(t-frac{pi}{N})psi(t)right|operatorname{d}t+ frac{1}{2} int_{-pi}^{pi}left|f_x(t-frac{pi}{N})psi(t)-f_x(t-frac{pi}{N})psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} int_{-pi}^{pi}left| f_x(t)-f_x(t-frac{pi}{N})right|operatorname{d}t+ frac{|f|_infty}{2} int_{-pi}^{pi}left|psi(t)-psi(t-frac{pi}{N}))right|operatorname{d}t \ le frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})$$
and since $f,psiin L^1([-pi,pi])$
$$sup_{xin[-pi,pi]}left|int_{[-pi,pi]backslash [-delta,delta]} f_x(t)psi(t)sin(Nt)frac{operatorname{d}t}{2pi}right|lefrac{1}{2pi}left(frac{|psi|_infty}{2} omega_{f,1}(frac{pi}{N})+ frac{|f|_infty}{2} omega_{psi,1}(frac{pi}{N})right)to 0, Nto+infty.$$
answered Jan 23 at 22:13
BobBob
1,7311725
edited Feb 3 at 0:23
answered Jan 23 at 22:13
BobBob
1,7311725
answered Jan 23 at 22:13
BobBob
1,7311725
answered Jan 23 at 22:13
BobBob
1,7311725
1,7311725
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$begingroup$
This is true and it is a basic theorem in the theory of Fourier series. Ref: Fourier Series by Edwards.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 10:28