$\delta_n(y-x) = \begin{cases} 0, & (y-x)\leq -\frac{1}{2n}\\ n, & -\frac{1}{2n}\leq (y-x) \leq \frac{1}{2n}\\ 0, & (y-x)\geq \frac{1}{2n} \end{cases}$ Note that _rectangle function_ is the common name given to normalized [[Boxcar function]]s. --- # Proofs and examples ## [Proof of convergence](Representation%20of%20𝛿(y-x).md#Proving%20that%20a%20function%20converges%20to%20𝛿%20y-x) to 𝛿(y-x) We can easily see the box function as given has an area of $1$ regardless of $n$ by multiplying the lengths of its sides, $(2/2n)n,$ thus proving that .md#^9ec0e7) without the need to integrate. In order to show .md#^58f03f) Plug in the function into the integral and change the limits to the domain of the box function, since we know by definition it's $0$ everywhere else. Thus, $\lim_{n\rightarrow\infty}\int_{-\infty}^{\infty} dx f(x) \delta_n(y-x)=\lim_{n\rightarrow\infty}\int_{-1/2n}^{1/2n} dx f(x)n$ $=\lim_{n\rightarrow\infty}\int_{-1/2n}^{1/2n} dx (f(y)+f'(y)(x-y)+\frac{1}{2}f''(y)(x-y)^2...)n=f(y)$ Simply taking the integral and then the limit gives the final answer: $\lim_{n\rightarrow\infty}\bigg(n(f(y)(x-y)+\frac{1}{2}f'(y)(x-y)^2+\frac{1}{3}\frac{1}{2}f''(y)(x-y)^3...)\bigg|_{-\frac{1}{2n}}^{\frac{1}{2n}}\bigg)$ $=\lim_{n\rightarrow\infty}\bigg(\frac{n}{n}f(y)+\frac{1}{2}f'(y)\frac{n}{2n^2}+\frac{1}{3}\frac{1}{2}f''(y)\frac{n}{4n^3}...\bigg) = f(y)$ where all terms except $f(y)$ disappear under the limit.  <font size="2"> The box function centered at $y=0$ converges to $\delta(x)$. The fact that the range of the function compresses to 0 is what allows us to prove this without integrating.</font> #MathematicalFoundations/Analysis/GeneralizedFunctions #MathematicalFoundations/Analysis/Functions