Vtgyjfy

Technically homework(a midterm) but its over and I'm itching to know the solution. I know how to show it for $alpha>1/2$ (the Fourier series will converge absolutely), but apparently its true for any $alpha$; the question guided me as follows:

1. Show that if a equicontinuous sequence of functions ($f_n$) converges pointwise to $f$, then $f_n$ converges uniformly to $f$.


2. Show for $f∈ C^alpha([0,1]/{sim})$ that $S_nf → f$ pointwise.


3. Show that the sequence $(S_nf)$ is equicontinuous and conclude.

1 and 2 posed no problems to me but I could not do 3. Any help? In addition, I would not mind other ways to prove the result.

Suppose that $|f(x)|le C$ and $|f(x)-f(y)|le C|x-y|^alpha$.

Express the Difference Using the Dirichlet Kernel

Using the Dirichlet Kernel, we get
$$
begin{align}
|S_nf(x)-f(x)|
&=left|,int_{-1/2}^{1/2}frac{sin((2n+1)pi y)}{sin(pi y)}[f(x-y)-f(x)],mathrm{d}y,right|\
&=left|,sum_{k=-n}^nint_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}frac{sin((2n+1)pi y)}{sin(pi y)}[f(x-y)-f(x)],mathrm{d}y,right|tag{1}
end{align}
$$

Estimate each Integral Using the smoothness of $boldsymbol{f}$

Since $left|,frac{sin((2n+1)pi y)}{sin(pi y)},right|lefrac{2n+1}{big|2|k|-1big|}$ and each interval is $frac1{2n+1}$ wide, we can bound
$$
begin{align}
left|,int_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}frac{sin((2n+1)pi y)}{sin(pi y)}[f(x-y)-f(x)],mathrm{d}y,right|
&lefrac{C}{big|2|k|-1big|}left(frac{2|k|+1}{4n+2}right)^alphatag{2}
end{align}
$$

Estimate each Integral Using Cancellation from $boldsymbol{sin((2n+1)pi x)}$

For $|y|lefrac12$, we have $|2y|le|sin(pi y)|le|pi y|$, and because
$$
int_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}sin((2n+1)pi y),mathrm{d}y=0tag{3}
$$
and
$$
int_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}|sin((2n+1)pi y)|,mathrm{d}y=frac2{(2n+1)pi}tag{4}
$$
if we let $m_k$ be the middle of the range of $frac{f(x-y)-f(x)}{sin(pi y)}$ on $left[frac{2k-1}{4n+2},frac{2k+1}{4n+2}right]$, for $kne0$, we can bound
$$
begin{align}
&left|,int_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}sin((2n+1)pi y)frac{f(x-y)-f(x)}{sin(pi y)},mathrm{d}y,right|\
&=left|,int_{frac{2k-1}{4n+2}}^{frac{2k+1}{4n+2}}sin((2n+1)pi y)left[frac{f(x-y)-f(x)}{sin(pi y)}-m_kright],mathrm{d}y,right|\
&lefrac1{(2n+1)pi}frac{overbrace{pifrac{2|k|+1}{4n+2}}^{sin(pi y)}overbrace{C(2n+1)^{-alpha}vphantom{frac{|}2}}^{Delta (f(x-y)-f(x))}+overbrace{2Cvphantom{()^1}}^{f(x-y)-f(x)}overbrace{pi(2n+1)^{-1}}^{Deltasin(pi y)}}{underbrace{frac{4k^2-1}{(2n+1)^2}}_{sin^2(pi y)}}\
&=frac{C(2n+1)^{-alpha}}{4|k|-2}+frac{2C}{4k^2-1}tag{5}
end{align}
$$

Use each Estimate in its Proper Place

If we use estimate $(2)$ for $kle m=n^{frac{alpha}{alpha+1}}$ and estimate $(5)$ for $kgt m$, then we get
$$
begin{align}
sum_{|k|le m}frac{C}{big|2|k|-1big|}left(frac{2|k|+1}{4n+2}right)^alpha
&lefrac{C}{(4n+2)^alpha}left[1+6sum_{k=1}^m(2k+1)^{alpha-1}right]\
&lefrac{C}{(4n+2)^alpha}frac3alpha(2m+1)^alpha\
&simfrac{3C}{alpha2^alpha}n^{-fracalpha{alpha+1}}tag{6}
end{align}
$$
and
$$
begin{align}
sum_{mlt|k|le n}frac{C(2n+1)^{-alpha}}{4|k|-2}
&lefrac{C}{2^{alpha+1}}frac{H_n}{n^alpha}\
&simfrac{C}{2^{alpha+1}}frac{log(n)}{n^alpha}\
&=oleft(n^{-frac{alpha}{alpha+1}}right)tag{7}
end{align}
$$
and
$$
begin{align}
sum_{mlt|k|le n}frac{2C}{4k^2-1}
&le Csum_{k=m}^inftyfrac1{k^2-1}\
&=frac{C}{2}sum_{k=m}^inftyleft(frac1{k-1}-frac1{k+1}right)\
&=frac{C}{2}left(frac1{m-1}+frac1mright)\
&sim Cn^{-frac{alpha}{alpha+1}}tag{8}
end{align}
$$

Put Everything Together

Therefore, we have uniform convergence:
$$
|S_nf(x)-f(x)|leleft(1+frac3{alpha2^alpha}right)Cn^{-frac{alpha}{alpha+1}}tag{9}
$$

While I accepted the answer above, this is how my lecturer (and later my friend) explained it to me (the exam is tomorrow). We first define $$g_n(x):=f(x) - S_n f(x)$$ just to remind ourselves that we need to be careful of cancellations. Then uniform convergence of $S_nf$ to $f$ is equivalent to showing $g_n→ 0$ uniformly; since we know (part 2) that $g_n(x) → 0$ pointwise, it suffices to show $g_n$ is uniformly continuous (by part 1).

Since $newcommand{d}{text{d}}newcommand{intT}{∫_{-1/2}^{1/2}}g_n(x) = f(x)times 1 - intT f(z-x) D_n(z) d z = intT [f(x) -f(z-x)] D_n(z) d z$ ,

begin{align}
|g_n(x) - g_n(y)| ≤ intT |D_n(z)|underbrace{|f(x) - f(z-x) - f(y) + f(z-y)|}_{(star)} d z
end{align}

We now need to find bounds independent of $n$ . We use a simple bound for the Dirichlet kernel $D_n$: as there is $C_0$ such that $|sin(2π z)|>C_0|z|$ on $[-1/2,1/2]$,
$$|D_n(z)| < frac{C_1}{|z|} $$

Since we don't gain too much form a simple bound we need to bound $(star)$. The trick is to use two different bounds, each good on different sets:

begin{align} |color{red}{f(x) - f(z-x)} - color{blue}{f(y) + f(z-y)}| &leq C_3|z|^alpha \
|color{red}{f(x)} - color{blue}{f(z-x)} - color{red}{f(y)} + color{blue}{f(z-y)}|
&leq C_3|x-y|^alpha \
end{align}

Thus $$|g_n(x) - g_n(y)| leq ∫_{|z|leq|x-y|} C_4|z|^{alpha-1} d z + |x-y|^alpha ∫_{|x-y|<|z|<1/2}frac{C_5}{z} d z = I_1 + I_2 $$

Now $I_1$ is $mathcal{O}(|x-y|)$ because $|z|^{alpha-1}$ is $L^1([-1/2,1/2])$. The second we compute,

$$I_2 = C_5 |x-y|^alphaleft(logfrac{1}{2} + logfrac{1}{|x-y|}right) $$
And we win because polynomials beat logarithms.

Actually I just used estimates in the same spirit of the ones used in the answer of Calvin Khor, proving directly the result without appelling to Ascoli-Arzelà theorem and obtaining also an estimate of the rate of convergence, so I think that it makes sense to post this answer.

First, get $f:mathbb{R}to mathbb{C}$ be a $2pi$-periodic $alpha$-Hölder continuous function and for each $xinmathbb{R}$ define $f_x(t):=f(x+t)-f(x)$.
We want to prove that:
$$sup_{xin[-pi,pi]}left|sum_{n=-N}^Nhat{f}(n)e^{inx}-f(x)right|to0, Nto+infty.$$
Now:
$$sum_{n=-N}^Nhat{f}(n)e^{inx}-f(x)=int_{-pi}^pi (f(x+t)-f(x))frac{sinleft((N+frac{1}{2})tright)}{sinleft(frac{t}{2}right)}frac{operatorname{d}t}{2pi}=int_{-pi}^pi f_x(t)frac{sinleft((N+frac{1}{2})tright)}{sinleft(frac{t}{2}right)}frac{operatorname{d}t}{2pi} \ = int_{-pi}^pi f_x(t)cotleft(frac{t}{2}right)sin(Nt)frac{operatorname{d}t}{2pi}+int_{-pi}^pi f_x(t)cosleft(Ntright)frac{operatorname{d}t}{2pi}= \ frac{1}{pi}int_{-pi}^pi frac{f_x(t)}{t}sin(Nt)operatorname{d}t+int_{-pi}^pi f_x(t)left(cotleft(frac{t}{2}right)-frac{2}{t}right)sin(Nt)frac{operatorname{d}t}{2pi}+int_{-pi}^pi f_x(t)cosleft(Ntright)frac{operatorname{d}t}{2pi}.$$

Now, the only problematic integral is the first, so let's estimate only this one.
We have that:

$$int_{-pi}^pi f_x(t)cotleft(frac{t}{2}right)sin(Nt)frac{operatorname{d}t}{2pi} = -int_{-pi}^pi f_x(t+frac{pi}{N})cotleft(frac{t+frac{pi}{N}}{2}right)sin(Nt)frac{operatorname{d}t}{2pi},$$
so:
$$left|int_{-pi}^pi f_x(t)cotleft(frac{t}{2}right)sin(Nt)frac{operatorname{d}t}{2pi}right| = left|frac{1}{2} int_{-pi}^pi left(f_x(t)cotleft(frac{t}{2}right)-f_x(t+frac{pi}{N})cotleft(frac{t+frac{pi}{N}}{2}right)right)sin(Nt)frac{operatorname{d}t}{2pi}right| \ le frac{1}{4pi} int_{-pi}^pi left|f_x(t)cotleft(frac{t}{2}right)-f_x(t+frac{pi}{N})cotleft(frac{t+frac{pi}{N}}{2}right)right|operatorname{d}t \ = frac{1}{4pi} int_{-pi}^pi left|f_x(t-frac{pi}{2N})cotleft(frac{t-frac{pi}{2N}}{2}right)-f_x(t+frac{pi}{2N})cotleft(frac{t+frac{pi}{2N}}{2}right)right|operatorname{d}t$$
So we need an uniform estimate in $x$ for the quantity:
$$int_{-pi}^pi left|f_x(t-h)cotleft(frac{t-h}{2}right)-f_x(t+h)cotleft(frac{t+h}{2}right)right|operatorname{d}t$$
for $hto 0$.

Split the integral for $|t|<2|h|$ and for $2|h|le|t|le pi$ for $|h|<1$. So:
$$int_{|t|<2|h|} left|f_x(t-h)cotleft(frac{t-h}{2}right)-f_x(t+h)cotleft(frac{t+h}{2}right)right|operatorname{d}t\ le int_{|t|<2|h|} left|f_x(t-h)cotleft(frac{t-h}{2}right)right|operatorname{d}t+int_{|t|<2|h|} left|f_x(t+h)cotleft(frac{t+h}{2}right)right|operatorname{d}t le \ int_{|t|<2|h|} |t+h|^alphaleft|cotleft(frac{t+h}{2}right)right|operatorname{d}t+int_{|t|<2|h|} |t-h|^alphaleft|cotleft(frac{t-h}{2}right)right|operatorname{d}t\ le c_1 int_{|t|<4|h|} |t|^{alpha-1}operatorname{d}t = c_2 |h|^alpha.$$
While, for $2|h|le|t|le pi$ we have:
$$int_{2|h|le|t|le pi} left|f_x(t-h)cotleft(frac{t-h}{2}right)-f_x(t+h)cotleft(frac{t+h}{2}right)right|operatorname{d}t \ le int_{2|h|le|t|le pi} left|f_x(t-h)left(cotleft(frac{t-h}{2}right)-frac{2}{t-h}right)right|operatorname{d}t+int_{2|h|le|t|le pi} left|f_x(t+h)left(cotleft(frac{t+h}{2}right)-frac{2}{t+h}right)right|operatorname{d}t+int_{2|h|le|t|le pi} left|f_x(t-h)frac{2}{t-h}-f_x(t+h)frac{2}{t+h}right|operatorname{d}t.$$
Now it is clear that the first two integrals aren't a problem, while for the last:
$$int_{2|h|le|t|le pi} left|f_x(t-h)frac{1}{t-h}-f_x(t+h)frac{1}{t+h}right|operatorname{d}t \ le int_{2|h|le|t|le pi} |t|left|frac{f_x(t-h)-f_x(t+h)}{(t-h)(t+h)}right|operatorname{d}t + |h|^2int_{2|h|le|t|le pi} left|frac{f_x(t-h)+f_x(t+h)}{(t-h)(t+h)}right|operatorname{d}t \ le |2h|^alpha int_{2|h|le|t|le pi} left|frac{t}{(t-h)(t+h)}right|operatorname{d}t + |h|int_{2|h|le|t|le pi} left|t-hright|^{alpha-1}operatorname{d}t+|h|int_{2|h|le|t|le pi} left|t+hright|^{alpha-1}operatorname{d}t \ le c_3|h|^{alpha}(1+log|h|).$$

Putting all together we get:
$$sup_{xin[-pi,pi]}left|sum_{n=-N}^Nhat{f}(n)e^{inx}-f(x)right|le Cleft|frac{pi}{N}right|^{alpha}log left|frac{pi}{N}right|to 0, Nto+infty$$