Showing Holder condition implies uniform convergence of the Fourier seires Suppose $f:\mathbb T\rightarrow \mathbb C$ satisfies $$|f(s)-f(t)|<M|s-t|^a$$ for every $s,t\in \mathbb T$ with $M>0$ and $ a\in (0,1]$ fixed.
I know that by the Riemann-Lebesgue Lemma, $$f(x)=\lim_{n \to \infty} s_n(f;x)$$ where $$s_n(f;x)=\int_{\mathbb T}f(x-t)\frac{\sin(n+\frac{1}{2})t}{\sin\frac{1}{2}t}dt$$
Furthermore, it is known that the convergence is uniform. And I am having trouble dealing with this uniform convergence part. I guess I should apply the Riemann-Lebesgue Lemma to some function $h(t)$ that uniformly bounds $$\frac{f(x-t)-f(x)}{\sin t/2}$$ in some appropriate way. But I cannot see it clearly. Can you help me with this? Thanks and regards.
 A: You are correct in thinking that uniformity of the Riemann-Lebesgue lemma is a key piece in this. And this does generally hold if you can approximate the function $f$ in question by a smooth function in the $L^1$ norm to arbitrary precision, which is always true if $f\in L^1$ or $f\in L^2$. Classical arguments work to approximate your Holder continuous function by a continuous piecewise linear function.
Let $D_N(x)=\frac{1}{\pi}\frac{\sin((n+1/2)x)}{\sin(x/2)}$, just to avoid normalization issues. Then the trucated Fourier series is
\begin{align}
                  S_N(f)(x) &= \int_{-\pi}^{\pi}D_N(y)f(x-y)dy \\
   &= \int_{0}^{\pi}D_N(y)\frac{f(x+y)+f(x-y)}{2}dy.
\end{align}
Suppose $f$ is as you describe, and let $M$ be a uniform bound for $f$.
Let $\delta \in (0,\pi)$ be fixed and small. Then there is a constant $K$ such that $1/\sin(x/2)$ is uniformly bounded by $K$ on $[\delta,\pi]$, and there is a piecewise linear continuous and $2\pi$ periodic function $f_{\epsilon}$ that is bounded by $M$ and satisfies
$$
                \int_{0}^{2\pi}|f(x)-f_{\epsilon}(x)|dx < \frac{\pi\epsilon}{K}.
$$
Then,
$$
      \left|\int_{\delta}^{\pi}D_N(y)\frac{f(x+y)+f(x-y)-f_{\epsilon}(x+y)-f_{\epsilon}(x-y)}{2}dy\right| \\
   \le \frac{K}{2\pi}\int_{\delta}^{\pi}|f(x+y)-f_{\epsilon}(x+y)|+|f(x-y)-f_{\epsilon}(x-y)| dy \\
  <\frac{K}{2\pi}\cdot\frac{\pi\epsilon}{K}=\frac{\epsilon}{2}
$$
Using integration by parts, it is easy to show that the following converges uniformly in $x$ to $0$.
$$
          \int_{\delta}^{\pi} D_N(y)\frac{f_{\epsilon}(x+y)+f_{\epsilon}(x-y)}{2}dy
$$
Therefore, the following also converges to $0$ uniformly as $N\rightarrow\infty$ for any fixed $\delta > 0$:
$$
     \int_{\delta}^{\pi}D_N(y)\frac{f(x+y)+f(x-y)}{2}dy.
$$
Therefore, for your function
$$
    S_N(f)(x)-f(x) = \int_{0}^{\delta}D_N(y)\left[\frac{f(x-y)+f(x+y)}{2}-f(x)\right]dy+E_N(x),
$$
where $E_N(x)$ converges uniformly to $0$ as $N\rightarrow\infty$, regardless of the choice of fixed $\delta > 0$. Finally, given $\epsilon > 0$, one can choose $\delta$ small enough that 
$$
     M\int_{0}^{\delta}\frac{|\sin((n+1/2)y|}{|\sin(y/2)|}|y|^{\alpha}dy < \frac{\epsilon}{2},
$$
and then choose $N_0$ large enough that $E_N(x)$ is uniformly bounded by $\frac{\epsilon}{2}$ for $N \ge N_0$. It then follows that $|S_N(f)(x)-f(x)| < \epsilon$ holds uniformly for all $x$. Hence, the Fourier series for $f$ converges uniformly to $f$.
A: I am assuming the following first:
$f \in L^{1}(\mathbb T) $, because what I think you are working with is Dini's condition for pointwise convergence on $L^{1}(\mathbb T)$, where $L^{1}(\mathbb T)$ is the class of all measurable functions $f$ defined a.e. on $\mathbb R$, such that $f$ has period $2 \pi$.  Now since $f$ satisfies the Holder's condition on all $t \in \mathbb T$, and we know that a is function $f$ satisfies Dini's condition at some point $t$, then partial sums converge. In particular, if there exists a $\delta > 0$ such that
$$\int_{0}^{\delta} \frac{|f(t+s) + f(t-s) - 2f(t)|}{s} ds < \infty$$
then $\lim_{n \to\infty} S_{N}f(t) = f(t)$.
Now note that, $f$ also satisfies Holder's condition for each $t$ i.e., $\exists M > 0$ such that
$$|f(t+s)-f(t)| \le M|s|^{a}, \quad \forall s, t$$ 
Therefore,
$$\int_{0}^{\delta} \frac{|f(t+s) + f(t-s) - 2f(t)|}{s} ds \le \int_{0}^{\delta} 2M |s|^{a-1} ds < \infty$$
And since this will hold true independently of $t$, the limit is uniform. 
