----- > [!proposition] Proposition. ([[Fourier series is Cesaro summable at points of continuity]]) > 1. If $f$ is [[Riemann integral|integrable]] [[function on the (unit) circle|on the circle]], then the [[Fourier series]] of $f$ is [[Cesaro summable]] to $f$ at every [[continuous]] point of $f$. Meaning, $\lim_{ N \to \infty } \sigma_{N}(f)(\theta_{0})=f(\theta_{0}).$ > 2. If $f$ is [[continuous]] [[function on the (unit) circle|on the circle]], then the [[Fourier series]] of $f$ is [[Cesaro summable]] to $f$ [[uniform convergence|uniformly]]. > 3. **Corollary 1.** If $f$ is [[Riemann integral|integrable]] [[function on the (unit) circle|on the circle]] with $\hat{f}(n) \equiv 0$, then $f=0$ at every point of continuity of $f$. [[uniqueness condition for Fourier Series|(compare to this!)]] > 4. **Corollary 2.** If $f$ is [[continuous]] [[function on the (unit) circle|on the circle]], then $f$ can be uniformly approximated by [[trigonometric polynomial]]s. So, the [[trigonometric polynomial]]s are dense in the space of [[continuous]] functions [[function on the (unit) circle|on the circle]] in a [[uniform convergence|uniform]] sense. > [!proof]- Proof. ([[Fourier series is Cesaro summable at points of continuity]]) >1. Recall that by the **remark** in [[Cesaro summable]] we have that $\sigma_{N}(f)=f * F_{N}$, where $F_{N}$ denotes the $N^{th}$ [[Fejer Kernel]]. Suppose $f$ is [[continuous]] at $\theta_{0}$. So we need to show $\lim_{ N \to \infty } (f * F_{N})(\theta_{0})=f(\theta_{0}) .$ Since $F_{N}$ is a [[good kernel]], by [[limit of good kernels approximates the convolution identity, given continuity]] we obtain the result. > 2. If $f$ is actually [[continuous]] on the whole circle we still obtain the result from the second part of that theorem. >3. By **Remark 2** in [[Cesaro summable|here]] we get that $\sigma_{N}(f)(\theta)$ is a [[linear combination]] of $\hat{ f}(k)$, $k=-N+1,\dots,N-1$, since by assumption each $\hat{f}(k)$ is $0$ we get too that each $\sigma_{N}(f)(\theta)$ is $0$. >4. Put $p_{N}(\theta)=\sigma_{N}(f)(\theta)$. Then $p_{N}$ is a [[trigonometric polynomial]] in $\theta$ and by $(2)$ we get that $p_{N}(\theta) \to f(\theta)$ [[uniform convergence|uniformly]] in $\theta.$ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```