# 小胖的故事

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• 红树

## 概要：

\begin{split}
\int_{y-\epsilon}^{y+\epsilon} \delta^{(2)}(x-y) f(x) dx &= f^{(2)}(y)
\end{split}

where ##\epsilon > 0##

Is the following also true as ##\epsilon \rightarrow 0##

\begin{split}
\int_{y-\epsilon}^{y+\epsilon} \delta^{(2)}(x-y) f(x) dx &\rightarrow \delta^{(2)}(x-y) f(x)
\\
&= f^{(2)}(y)
\end{split}

## 答案与答复

The integral is constant for ##\epsilon > 0##, hence the limit as ##\epsilon \rightarrow 0^+## is ##f(y)##.

I'm not sure that the exponent ##^{(2)}## means.

##\delta^{(2)}(x)= \delta''(x)##小胖的故事

My question: ##\delta''(x-y) f(y) \stackrel{?}{=} f''(y)##

##\delta^{(2)}(x)= \delta''(x)##
The same applies. If it has that value for all ##\epsilon > 0## then the right-hand limit exists and is equal to the constant value. That doesn't, however, mean that you can set ##\epsilon = 0##.

\begin{split}
\delta(x-y) f(x) &= \begin{array}{cc}
f(y) & \text{ if } x=y \\
0 & \text{ if } x \neq y
\end{array}
\end{split}

It is simply a simplification of this notation to write ##\delta(x-y) f(x)=f(y)##

\begin{split}
f(y) &= \delta(x-y) f(x)
\end{split}

##\delta^{(2)}(x)= \delta''(x)##

My question: ##\delta''(x-y) f(y) \stackrel{?}{=} f''(y)##
That would imply that (for all ##x, y## and functions ##f##) we have ##\delta''(x-y)= \frac{f''(y)}{f(y)}##.

It is true that ##\delta''(x) * f(x) = \int_{y-\epsilon}^{y+\epsilon} \delta''(x-y) f(x) dx = f''(y)##

\begin{split}
\delta''(x-y) f(x) = \begin{array}{cc}
f''(y) & \text{ if } x=y \\
0 & \text{ if } x\neq y
\end{array}
\end{split}

which can be simplified to ##\delta''(x-y) f(x) = f''(y)##

It is true that ##\delta''(x) * f(x) = \int_{y-\epsilon}^{y+\epsilon} \delta''(x-y) f(x) dx = f''(y)##

\begin{split}
\delta''(x-y) f(x) = \begin{array}{cc}
f''(y) & \text{ if } x=y \\
0 & \text{ if } x\neq y
\end{array}
\end{split}

which can be simplified to ##\delta''(x-y) f(x) = f''(y)##

You are right, except that ##x## and ##y## can be considered dummy variables in the convolution ##\delta'' * f = f''##

$$f(y) = \delta(x-y) f(x)$$

$$\int_{-\infty}^\infty \delta(x-y) f(x)\,dx = f(y).$$ Note that inclusion of the integral eliminates the problem @PeroK 注意。双方都是仅## y ##的功能。

$$\delta(x) = \begin{cases} +\infty & x = 0 \\ 0 & x \ne 0 \end{cases}$$ so
$$f(x)\delta(x-y) = \begin{cases} \operatorname{sgn}(f(y))\cdot \infty & x = y \\ 0 & x \ne 0 \end{cases}$$ Only by integrating do you get a finite result.