The proof of this is something students get tested on so let’s go over it here: \begin{proof} Let $\phi \in \mathcal{S}$. Then, $$\begin{align} \int\limits \psi_{n}(t)\phi(t) \, dt -\phi(0)&= \int\limits \psi_{n}(t)\phi(t) \, dt -\int\limits \psi_{n}(t)\phi(0) \, dt\\ &= \int\limits \psi_{n}(t)(\phi(t)-\phi(0)) \, dt \tag{😂} \end{align}$$ where the first equality uses the second property of s and the second applies linearity of the Riemann integral.

Since $\phi$ is continuous then $\forall\epsilon>0,\exists\delta_{\epsilon}>0$ such that $\forall|t|\le \delta_{\epsilon}$, we have that $|\phi(t)-\phi(0)| \le \epsilon$. Accordingly, $$\begin{align*} &\limsup_{ n \to \infty } \left| \int\limits \psi_{n}(t)\phi(t) \, dt-\phi(0) \right| = \limsup_{ n \to \infty } \left| \int\limits \psi_{n}(t)(\phi(t)-\phi(0)) \, dt \right| \\ &\le \limsup_{ n \to \infty } \left| \int\limits _{\{ t:|t|>\delta_{\epsilon} \}}\psi_{n}(t)(\phi(t)-\phi(0)) \, dt \right| + \limsup_{ n \to \infty } \left| \int\limits _{\{ t:|t|\le\delta_{\epsilon} \}}\psi_{n}(t)(\phi(t)-\phi(0)) \, dt \right| \\ &\le \limsup_{ n \to \infty }\int\limits _{\{ t:|t|>\delta_{\epsilon} \}}\psi_{n}(t) \left| \phi(t)-\phi(0) \right| \, dt + \limsup_{ n \to \infty }\int\limits _{\{ t:|t|\le\delta_{\epsilon} \}}\psi_{n}(t) \left| \phi(t)-\phi(0) \right| \, dt \\ &\le \left( 2\sup_{t\in \mathbb{R}}|\phi(t)| \right) \limsup_{ n \to \infty }\int\limits _{\{ t:|t|>\delta_{\epsilon} \}}\psi_{n}(t)\, dt + \limsup_{ n \to \infty }\epsilon\int\limits _{\{ t:|t|\le\delta_{\epsilon} \}}\psi_{n}(t) \, dt \\ &\le \epsilon \end{align*}$$ Where we:

1. Apply $(😂)$
2. Triangle inequality of the 1 norm.
3. Integral Triangle Inequality plus property 1 of
4. 1st term we take total variation bound (idk how this works rigorously) 2nd term we use dialogue before equations (i.e., use fact that $\phi$ is continuous and the definition of $\delta_{\epsilon}$).
5. 1st term is 0 by third property and for the 2nd term the integral is 1 by property 2

Since $\epsilon$ is arbitrary, the result follows so that the limit above exists and is zero hence $$\int\limits \psi_{n}(t)\phi(t) \, dt \to\bar{\delta}(\phi)=\phi(0)$$ \end{proof} >[!example] >An example of an is $$f_{n}(t)=\begin{cases} n & 0\le t\le \frac{1}{n}\\0 & \text{else}\end{cases}$$We can define the distribution $$\bar{f}_{n}(\phi):=\int\limits _{0}^{\infty}f_{n}(t)\phi(t) \, dt .$$

Recall the Convolution operation: Convolution

Where we can restrict $f,g$ to be in$L^{2}$$(\mathbb{R};\mathbb{R})$. We then have the following result

Now, using and we can prove the following: >[!thm|3.4.4] >The family of complex exponentials in $L^{2}$$([-\pi,\pi];\mathbb{C})$: $$\{ e_{n}(t) \}_{n\in \mathbb{Z}}=\left\{ \frac{1}{\sqrt{ 2\pi }}e^{int} \right\}_{n\in \mathbb{Z}}$$forms a complete orthonormal sequence.