[[Noteworthy Uses]]:: *[[Noteworthy Uses]]*
[[Proved By]]:: *[[Proved By|Crucial Dependencies]]*
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- Let $\{ k_{n}(x) \}_{n=1}^{\infty}$ be a family of [[good kernel]]s;
- Let $f$ be an [[Riemann integral|integrable]] [[function on the (unit) circle|on the circle]];
> [!theorem] Theorem. ([[limit of good kernels approximates the convolution identity, given continuity]])
> Then $\lim_{ n \to \infty }(f * k_{n})(x_{0})=f(x_{0})$ provided that $f$ is [[continuous]] at $x_{0}$.
> \
> If $f$ is [[continuous]] [[function on the (unit) circle|on the circle]], then the [[uniform convergence|convergence is in fact uniform]] in $x$.
> [!proof]- Proof. ([[limit of good kernels approximates the convolution identity, given continuity]])
>
> Let $f$ be [[continuous]] at $x_{0}$, fix $\varepsilon >0$, and define $\varepsilon':= \frac{\varepsilon}{2M}$. So, $\ex \delta>0$ s.t. $|f(s)-f(x_{0})|\leq \varepsilon'$ whenever $|s-x_{0}| \leq \delta$. Recalling the [[good kernel]] definition, we can write $\begin{align}
> (f * k_{n})(x_{0})-f(x_{0})= & (f * k_{n})(x_{0})-\frac{f(x_{0})}{2\pi}\int_{-\pi}^{\pi} k_{n}(y)\, dy \\
> = & \frac{1}{2\pi}\int_{-\pi}^{\pi}\big(f(x_{0}-y)-f(x_{0})\big)k_{n}(y)\,dy.
> \end{align}$
> [[modulus|Modulusing]] both sides we obtain $|(f * k_{n})(x_{0})-f(x_{0})|\leq \frac{1}{2\pi} \int_{-\pi}^{\pi}|f(x_{0}-y)||k_{n}(y)|\,dy . (*)$
> The next big idea is to split up this integral $(*)$ based on the third [[good kernel]] property; write $\begin{align}
> (*)= & \overbrace{\frac{1}{2\pi} \int_{|y|<\delta}^{}|f(x_{0}-y)||k_{n}(y)|\,dy }^{(a)} + \overbrace{\frac{1}{2\pi} \int_{\delta < |y|<\pi}^{}|f(x_{0}-y)||k_{n}(y)|\,dy }^{(b)}.
> \end{align}$
> Now, $\begin{align}
> (a)= & \frac{1}{2\pi} \int_{|y|<\delta}^{}|f(x_{0}-y)||k_{n}(y)|\,dy \\
> \leq & \frac{1}{2\pi} \int_{|y|<\delta}^{} \varepsilon' |k_{n}(y)| \, dy \\
> = & \frac{\varepsilon'}{2\pi}\int_{|y|<\delta}^{} \, |k_{n}(y)| \, dy \\
> \leq & \frac{\varepsilon'}{2\pi} \int_{-\pi}^{\pi} |k_{n}(y)|\, dy \\
> \leq & \varepsilon'M \\
> = & \frac{\varepsilon}{2}
> \end{align}$
> and this holds $\fa n$.
> \
> Next, $f$ is [[Riemann integral|Riemann integrable]] [[function on the (unit) circle|on the circle]], hence [[bounded function|bounded]] by some $B>0$. Using this we have $\begin{align}
> (b) \leq & \frac{1}{2\pi} \int_{\delta \leq |y| \leq \pi}^{} | k_{n}(y) |\, dy \\
> = & \frac{B}{\pi} \int_{\delta \leq |y| \leq \pi}^{} |k_{n}(y)|\, dy \\
> \leq & \frac{\varepsilon}{2} \text{ when }n \text{ is big enough, by 3rd GK prop.}
> \end{align}$
> Hence there exists $N \in \nn$ s.t. $|(f * k_{n})(x_{0})-f(x_{0})| < \varepsilon$ for all $n \geq N$, since $|(f * k_{n})(x_{0})-f(x_{0})| \leq (a)+(b) < \frac{\varepsilon}{2}+\frac{\varepsilon}{2}$.
>
> For the second claim: if $f$ is [[continuous]] [[function on the (unit) circle|on the circle]], then $\delta$ may be chosen independently of $x_{0}$, thus $N$ can be chosen independently of $x_{0}$, thus we have [[uniform convergence]] (check.)
> [!intuition]
> Think Dirac Delta!
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#### References
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