I need help proving a theorem stated during a lecture of the Fuorier's Analysis course.
A Riemann-integrable function $f : \mathbb{R} \rightarrow \mathbb{C}$ is said to satifies the uniform Dini condition (UDC) on a compact set $A \subset \mathbb{R}$ if $\forall$$\epsilon >0$ $\exists \delta = \delta(\epsilon) > 0$ such that
$$\sup_{x \in A} \int_{-\delta}^\delta \Big|\frac{f(x + y) – f(x)}{y}\Big|dy < \epsilon$$
Theorem: Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be Riemann-integrable and $2\pi$-periodic. If $f$ satisfies UDC on $[-\pi,\pi]$ then $S_N^f \to f$ uniformly on $[-\pi,\pi]$.
Here $S_N^f(x) := \sum_{n = -N}^N\hat{f}(n)e^{inx}$ and $\hat{f}(n) := \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}dx$. The identities $S_N^f(x) = \frac{1}{2\pi}\int_{-\pi}^\pi D_N(y)f(x – y)dy$ and $\frac{1}{2\pi}\int_{-\pi}^\pi D_N(x)dx = 1$ could be useful ($D_N(x) := \sum_{n = -N}^Ne^{inx}$ is the Dirichlet kernel)
My attempt: I know from a previuos theorem that $S_N^f(x) \to f(x)$ pointwise for every $x \in [-\pi,\pi]$. I tried to prove that $(S_N^f)_N$ is uniformly Cauchy: let $M > N$
$$\Big|\sum_{-M}^M \hat{f}(n)e^{inx} – \sum_{-N}^N \hat{f}(n)e^{inx}\Big| = \Big|\sum_{N + 1}^M(\hat{f}(n) + \hat{f}(-n))\Big| \leq \sum_{N + 1}^M|\hat{f}(n) + \hat{f}(-n)|$$
I think that I have to use the UDC in some way into the integral
$$\hat{f}(n) + \hat{f}(-n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(y)(e^{-iny} + ^{iny})dy$$
but I cannot see how. Please help me, any hint will be appreciated very much
EDIT: following @runs's suggestion I get the following (I suppose $f$ to be $\mathbb{R}$-valued)
\begin{align*}
|S_N^f(x) – f(x)| & = \frac{1}{2\pi}\Big|\int_{-\pi}^\pi D_N(y)(f(x – y) – f(x))dy\Big|\\
& = \frac{1}{2\pi}\Big|\int_{-\pi}^\pi D_N(t)(f(x + t) – f(x))dt\Big|
\end{align*}
now exploit the UDC
$$\leq \frac{1}{2\pi}\int_{-\delta}^\delta |D_N(t)t|\Big|\frac{f(x + t) – f(x)}{t}\Big|dt + \frac{1}{2\pi}\Big|\int_{I_\delta} D_N(t)(f(x + t) – f(x))dt\Big|$$
where $I_\delta := [-\pi,-\delta)\cup(\delta,\pi]$
$$K := \sup_{t \in [-\pi,\pi]}|D_N(t)t| < + \infty$$
so for the first integral we have
$$\frac{1}{2\pi}\int_{-\delta}^\delta |D_N(t)t|\Big|\frac{f(x + t) – f(x)}{t}\Big|dt \leq \frac{K}{2 \pi}\epsilon \:\:\: \text{for every}\: x \in [-\pi,\pi]$$
the map
$$[-\pi,\pi] \times I_\delta \ni (x,t) \mapsto f(x + t) – f(x)$$ is bounded so if we define
$$L := \sup_{[-\pi,\pi] \times I_\delta}\{f(x + t) – f(x)\}$$
$$l := \inf_{[-\pi,\pi] \times I_\delta}\{f(x + t) – f(x)\}$$
in the second integral we have
$$\frac{l}{2 \pi}\int_{I_\delta}D_N(t)dt \leq \frac{1}{2\pi}\int_{I_\delta} D_N(t)(f(x + t) – f(x))dt \leq \frac{L}{2 \pi}\int_{I_\delta}D_N(t)dt \:\:\: \text{for every}\: x \in [-\pi,\pi]$$
To conclude it suffices to prove that
$$\int_{I_\delta}D_N(t)dt \to 0 \:\:\:\text{as} \:\: N \to \infty \:\:\:\text{for every fixed}\: \delta$$
Let's prove this:
$$\int_{I_\delta}D_N(t)dt = \int_{-\pi}^\pi \chi_{I_\delta}(t) D_n(t)dt = \int_{-\pi}^\pi \frac{\chi_{I_\delta}(t)}{\sin(\frac{t}{2})}\sin{((N + \frac{1}{2})t)}dt$$
since $g(t) := \frac{\chi_{I_\delta}(t)}{\sin(\frac{t}{2})} \in L^1(-\pi,\pi)$ we can conclude by the Riemann-Lebesgue lemma.
Is my argument correct?
Best Answer
When $f$ is continuous,
Set $$g_\delta(x,y)= \frac{f(x+y)-f(x)}{\sin(y/2)} 1_{|\sin(y/2)|>\delta}$$ which is bounded, $4\pi$ periodic, uniformly continuous away from the neighborhood of the jump at $\sin(y/2)=\pm \delta$, so that $$\lim_{n\to \infty} \sup_x|\int_{-2\pi}^{2\pi} g_\delta(x,y)\sin((n+1/2)y)dy|$$ $$=\lim_{n\to \infty} \sup_x\frac12 |\int_{-2\pi}^{2\pi} (g_\delta(x,y)-g_\delta(x,y+\pi/n))\sin((n+1/2)y)dy|=0 $$ From there you get that for all $\delta>0$ $$\lim_{n\to \infty}\sup_x |f(x)-f\ast D_n(x)|= \lim_{n\to \infty}\sup_x |\int_{-\delta}^\delta \frac{f(x+y)-f(x)}{2\pi\sin(y/2)} \sin((n+1/2)y)dy|$$ And the RHS can be made arbitrary small by your uniform Dini criterion so that the LHS converges to $0$.