How to show for $\alpha\in (0,1)$, any $f\in C^\alpha([0,1]/{\sim})$ has a Fourier series $S_nf$ uniformly converging to $f$

Question

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.

Answer 1

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

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Express the Difference Using the Dirichlet Kernel

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

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

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

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

For $|y|\le\frac12$, we have $|2y|\le|\sin(\pi y)|\le|\pi y|$, and because $$ \int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\,\mathrm{d}y=0\tag{3} $$ 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 $k\ne0$, we can bound $$ \begin{align} &\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\frac{f(x-y)-f(x)}{\sin(\pi y)}\,\mathrm{d}y\,\right|\\ &=\left|\,\int_{\frac{2k-1}{4n+2}}^{\frac{2k+1}{4n+2}}\sin((2n+1)\pi y)\left[\frac{f(x-y)-f(x)}{\sin(\pi y)}-m_k\right]\,\mathrm{d}y\,\right|\\ &\le\frac1{(2n+1)\pi}\frac{\overbrace{\pi\frac{2|k|+1}{4n+2}}^{\sin(\pi y)}\overbrace{C(2n+1)^{-\alpha}\vphantom{\frac{|}2}}^{\Delta (f(x-y)-f(x))}+\overbrace{2C\vphantom{()^1}}^{f(x-y)-f(x)}\overbrace{\pi(2n+1)^{-1}}^{\Delta\sin(\pi y)}}{\underbrace{\frac{4k^2-1}{(2n+1)^2}}_{\sin^2(\pi y)}}\\ &=\frac{C(2n+1)^{-\alpha}}{4|k|-2}+\frac{2C}{4k^2-1}\tag{5} \end{align} $$

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Use each Estimate in its Proper Place

If we use estimate $(2)$ for $k\le m=n^{\frac{\alpha}{\alpha+1}}$ and estimate $(5)$ for $k\gt m$, then we get $$ \begin{align} \sum_{|k|\le m}\frac{C}{\big|2|k|-1\big|}\left(\frac{2|k|+1}{4n+2}\right)^\alpha &\le\frac{C}{(4n+2)^\alpha}\left[1+6\sum_{k=1}^m(2k+1)^{\alpha-1}\right]\\ &\le\frac{C}{(4n+2)^\alpha}\frac3\alpha(2m+1)^\alpha\\ &\sim\frac{3C}{\alpha2^\alpha}n^{-\frac\alpha{\alpha+1}}\tag{6} \end{align} $$ and $$ \begin{align} \sum_{m\lt|k|\le n}\frac{C(2n+1)^{-\alpha}}{4|k|-2} &\le\frac{C}{2^{\alpha+1}}\frac{H_n}{n^\alpha}\\ &\sim\frac{C}{2^{\alpha+1}}\frac{\log(n)}{n^\alpha}\\ &=o\left(n^{-\frac{\alpha}{\alpha+1}}\right)\tag{7} \end{align} $$ and $$ \begin{align} \sum_{m\lt|k|\le n}\frac{2C}{4k^2-1} &\le C\sum_{k=m}^\infty\frac1{k^2-1}\\ &=\frac{C}{2}\sum_{k=m}^\infty\left(\frac1{k-1}-\frac1{k+1}\right)\\ &=\frac{C}{2}\left(\frac1{m-1}+\frac1m\right)\\ &\sim Cn^{-\frac{\alpha}{\alpha+1}}\tag{8} \end{align} $$

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Put Everything Together

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

Answer 2

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,

$$I_2 = C_5 |x-y|^\alpha\left(\log\frac{1}{2} + log\frac{1}{|x-y|}\right) $$ And we win because polynomials beat logarithms.