[[Noteworthy Uses]]:: *[[Noteworthy Uses]]*
[[Proved By]]:: *[[Proved By|Crucial Dependencies]]*
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> [!theorem] Theorem. ([[smoother functions' Fourier coefficients decay faster]])
> Let $k \geq 1$. If $f$ is $C ^{k}$ [[function on the (unit) circle|on the circle]], then $\ex B>0$ s.t. $|\hat{f}(n)| \leq \frac{B}{|n|^{k}}, \ \ \fa n \neq 0.$
^edaea1
> [!intuition]
> We often think about 'smoother' signals as possessing 'less high frequencies'. For example,
> - [[Convolution]] of images $f$ with [[Gaussian]] kernels $g$ is used to model the image blurring ('smoothing') process, as each pixel is replaced with a weighted average of its neighbors. A [[Gaussian]] in the time domain corresponds to a gaussian in the frequency domain — a [[low-pass filter]] in light of [[the convolution theorem for Fourier series|the convolution theorem]] — and hence we intuit that smoother signals such as $f * g$ have less high frequencies than 'less smooth' signals such as $f$.
> - More generally, [[Convolution creates continuous functions]]; moreover, if $f \in C^{k}$ and $g \in C^{r}$ then $f * g \in C^{\max(k,r)}$. So in the interesting cases, convolution with a low-pass filter (e.g., [[rectangle]], [[Gaussian]], etc) removing high frequencies is done via convolution with a low-pass filter, which incidentally smoothens your function too.
> - One often images a prototypical 'noisy signal' as one which 'jitters about' some apparent [[ground-truth]] signal. The 'jittery-ness' of this signal intuitively gives the impression that lots of high frequencies are present... lots of 'sines would be needed' to approximate it
> - When approximating certain discontinuous functions, such as the [[Fourier series of a square wave]], [[Gibbs phenomenon]] shows that no matter how many terms you add together, the approximation will struggle heavily at the discontinuities.
> - In fact, the approximation will struggle *everywhere* — we see that isolated discontinuities — a *local* phenomenon — affect the *global* approximation.
> - This is in part when we need to introduce the notion of [[L2 convergence of Fourier Series|mean-square convergece of Fourier series]] for functions not sufficiently smooth.
> - The [[partial sum]]s of the [[Dirichlet Kernel]], in a loose sense, appear to approach $\delta$. In a sense, this is a [[Fourier series|Fourier partial sum]] in which all coefficients are $1$, illustrating that to approximate a not-very-smooth function with a 'discontinuity' as 'dramatic' as $\delta$ at $0$, we need all the high frequencies we can get.
>
> None of these examples provide anything more than a correlation between smoother functions and less high frequencies, but they do hint at a relationship which we are able to precisely state here.
^4765de
> [!proof]- Proof. ([[smoother functions' Fourier coefficients decay faster]])
> Recall that [[derivative of Fourier coefficient of C1 function]]. From this we ascertain that for the $C^k$ $f$ assumed we have $\hat{f}^{(k)}(n)=(in)^{k} \hat{f}(n)$. Thus $\hat{f}(n)=\frac{\hat{f} ^{(k)}(n)}{(in) ^{k}}, \ \ \fa n \neq 0$. $(*)$
\
Since $f$ is $C^k$ [[function on the (unit) circle|on the circle]], $f ^{(k)}$ is [[continuous]] [[function on the (unit) circle|on the circle]], hence [[continuous]] on $[-\pi,\pi]$, hence [[bounded function|bounded]] ([[continuous]] on [[compact]]), hence there exists $B>0$ s.t. $|f ^{(k)}(\theta)| \leq B$ for all $\theta$. Thus by [[bounded function has bounded Fourier coefficients]] we get that $|\hat{f} ^{(k)}(n)| \leq B$ for all $n \in \zz$. Now take the [[modulus]] on both sides of $(*)$ to obtain $|\hat{f}(n)|= |\frac{\hat{f} ^{(k)}(n)}{n ^{k}}| \leq \frac{B}{|n |^k}$
as desired.
^f581a4
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####
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#### 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
> ```