[[Noteworthy Uses]]:: *[[Noteworthy Uses]]* [[Proved By]]:: *[[Proved By|Crucial Dependencies]]* ---- - Let $f$ be [[Riemann integral|Riemann integrable]] [[function on the (unit) circle|on the circle]] with $\hat{f}(n)=0$ for all $n \in \zz$. > [!theorem] Theorem. ([[uniqueness condition for Fourier Series]]) > If $f$ is [[continuous]] at $\theta _0$, then $f(\theta _{0})=0$. > [!proposition] Corollary. > If $f$ is [[continuous]] [[function on the (unit) circle|on the circle]] and $\hat{f}(n) \equiv 0$, then $f \equiv 0$. > **Remark.** (From this it follows that if $f,g$ are [[continuous]] with the same [[Fourier series|Fourier coefficient]]s, then $f=g$.) > [!proof]- Proof of Statement. ([uniqueness condition for Fourier Series](app://obsidian.md/uniqueness%20condition%20for%20Fourier%20Series)) > We'll first show that if the theorem holds for all real-valued functions, then it holds for all complex-valued functions. Then we'll prove the theorem for real-valued functions. > **Step 1.** > Let $f(\theta):=u(\theta)+iv(\theta)$ satisfy $\hat{f}(n)\equiv 0$. Let $\theta_{0}$ be a point of [[continuous|continuity]] of $f$. We first claim that this implies $\hat{u}(n) \equiv_{}0$ and $\hat{v}(n) \equiv 0$. To see this, note that $\overline{f(\theta)}=u(\theta)-iv(\theta)$. Thus, $u(\theta)=\frac{1}{2}\big(f(\theta)+\overline{f(\theta)}\big)$, implying $\hat{u}(n)=\frac{1}{2}\big(\hat{f}(n) + \hat{\overline{f}}(n)\big)$. By the [[conjugate symmetry property of Fourier series]] we must then have $\hat{u}(n)= \frac{1}{2} \big(\hat{f}(n)+ \hat{f}(-n)\big)=\frac{1}{2}(0+0)=0.$ > An analogous computation shows $\hat{v}(n)=0.$**Step 2**. Now assume $f$ is real-valued with $\hat{f}(n) \equiv 0$ and let $\theta_{0}$ be a point of [[continuous|continuity]] of $f$. WLOG assume $\theta_{0}=0$ and $c:=f(\theta_{0})>0$. BWOC assume $f(\theta_{0}) \neq 0$; the idea is then to construct a family of [[trigonometric polynomial]]s $\{ p_{k} \}$ that "peak" at $\theta_{0}$ and so that $\int p_{k} (\theta) f(\theta) \, d\theta \to \infty$ as $k \to \infty$. We will see that, on the one hand, $\int p_{k} (\theta) f(\theta) \, d\theta =0 \ \ \fa k$ for our chosen $\{ p_{k} \}$, while on the other hand for large enough $k$ $\int p_{k} (\theta) f(\theta) \, d\theta >0$, a contradiction. > \ > Since $f$ is [[continuous]] at $0$, we can choose $0<\delta < \frac{\pi}{2}$ such that $f(\theta) \geq \frac{c}{2}$ for all $|\theta|<\delta$. [^1] Let $\varepsilon := \frac{2}{3}(1- \cos \delta)>0$, then define $p(\theta):= \varepsilon + \cos \theta$. For $\delta \leq |\theta| \leq \pi$, $p(\theta)\leq p(\delta)=\varepsilon + \cos \delta = \varepsilon + \left( 1-\frac{3}{2} \varepsilon \right) = 1- \frac{\varepsilon}{2}. $ > ![[CleanShot 2023-03-31 at [email protected]|400]] > Since $p(0)=1+\varepsilon > 1$ and $p$ is [[continuous]] at $0$, there exists $\eta \in (0, \delta)$ s.t. $p(\theta) \geq 1+ \frac{\varepsilon}{2}$ if $|\theta|<\eta$. (should be clear from picture) > > For each $k \in \nn$, define $p_{k}(\theta):=p(\theta) ^{k}=\varepsilon+ (\frac{e ^{i \theta}+ e ^{-i \theta}}{2}) ^{k}$, which is a [[linear combination]] of > [^1]: Recalling that $c=f(\theta_{0})=f(0)$, we can choose $0<\delta$ such that $| f(\theta)-f(c) | < \varepsilon$ whenever $| \theta - 0 |< \delta$ for any choice of $\varepsilon >0$. Pick $\varepsilon := \frac{c}{2}$. Now the [[absolute value]] implies that $f(\theta)-c < \frac{c}{2}$ and $c-f(\theta) < \frac{c}{2}$. We care about the second inequality, as rearranging it yields $f(\theta)> \frac{c}{2}$. Finally, we may make $\delta$ as small as we'd like now that we have it, so make it smaller than $\frac{\pi}{2}$. > [!proof]- Proof of Corollary. (Trivial) > Suppose $f$ is [[continuous]] [[function on the (unit) circle|on the circle]]. Let $\theta \in \dom f$. By [[uniqueness condition for Fourier Series]] we have that $f(\theta)=0$ since $f$ is by assumption [[continuous]] at $\theta$. Since $\theta$ was arbitrary we're done. > [!basicnonexample] Warning. > Weird things can happen without the [[continuous|continuity]] condition. An example that is easy to see would be letting $f(\theta)$ equal $0$ at all but a single point. Then its [[Fourier series|Fourier coefficient]]s will all be $0$ but $f(\text{that point})$ will not be! ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as 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 > ```