Verify the operator equation $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{sgn}\mathit{x}\mathbf{=}\mathbf{2}\mathit{\delta }\mathbf{\left(}\mathit{x}\mathbf{\right)}$where the function, meaning "sign ofx" and abbreviated, is defined by

$$\begin{gathered} \mathbf{sgn}\mathit{x}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{>}\mathbf{0}\mathbf{,} \\ \mathbf{-}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{<}\mathbf{0}\mathbf{\}} \end{gathered}$$

The given expressions are $\frac{d}{dx}sgnx=2\delta \left(x\right)$.

A function operator is a function that takes one (or more) functions as input and returns a function as output. In some ways, function operators are similar to functional

Let $G_1\left(x\right)=\frac{d}{dx}sgnx$and $G_2\left(x\right)=2\delta \left(x\right)$.

To verify the operator equation

$G_1\left(x\right)=G_2\left(x\right)$,

show that the

$\int \varphi \left(x\right)G_1\left(x\right)dx=\int \varphi \left(x\right)G_2\left(x\right)dx$

For any arbitrary test function $\varphi \left(x\right)$which is zero for $\left|x\right|>R$for some finite $R$.

Assuming $\varphi \left(x\right)$is such a decent test function, and then do the (integrating by parts)

$$\begin{gathered} \int \varphi \left(x\right)\frac{d}{dx}sgnxdx=\left(x\right)sgnx\int -\int sgnx\frac{d}{dx}\varphi \left(x\right)dx \\ ={\int }_{-\infty }^{\infty }\frac{d\varphi }{dx}dx-\underset{-\infty }{\overset{\infty }{\int }}\frac{d\varphi }{dx}dx \\ =\varphi \left(0\right)+\varphi \left(0\right) \\ =2\varphi \left(0\right) \end{gathered}$$

At the same time, the property of the delta function implies that,

$\int \varphi \left(x\right)2\delta \left(x\right)dx=2\varphi \left(0\right)$

Therefore, it is the case that $\int \varphi \left(x\right)\frac{d}{dx}\sin xdx=\int \varphi \left(x\right)2\delta \left(x\right)dx$for any test function $\varphi \left(x\right)$, which means that $\frac{d}{dx}\sin x=2\delta \left(x\right)$

It is shown in figure which is given below