Integrating with the Dirac delta distribution

[redtree]

TL;DR

In the definite integral of a delta function, how narrow can the interval be?

Given
\begin{equation}
\begin{split}
\int_{y-\epsilon}^{y+\epsilon} \delta^{(2)}(x-y) f(x) dx &= f^{(2)}(y)
\end{split}
\end{equation}
where $\epsilon > 0$

Is the following also true as $\epsilon \rightarrow 0$
\begin{equation}
\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}
\end{equation}

If not, why?

 

[PeroK]

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.

 

[redtree]

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

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

 

[PeroK]

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

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 $\epsilon = 0$.

 

[redtree]

For any value of x other than y, the equation is zero. The equation can be written as follows:
\begin{equation}
\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}
\end{equation}

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

Frankly, any function can be written in this fashion (and the same for any derivative), such that
\begin{equation}
\begin{split}
f(y) &= \delta(x-y) f(x)
\end{split}
\end{equation}

I don't see why the integral is necessary.

 

[PeroK]

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

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

That would imply that (for all $x, y$ and functions $f$) we have $\delta''(x-y)= \frac{f''(y)}{f(y)}$.

Which can't be right at all.

 

[redtree]

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

Thus, why would it not also be true that:
\begin{equation}
\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}
\end{equation}
which can be simplified to $\delta''(x-y) f(x) = f''(y)$

 

[PeroK]

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

Thus, why would it not also be true that:
\begin{equation}
\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}
\end{equation}
which can be simplified to $\delta''(x-y) f(x) = f''(y)$

It makes no sense. You can't equate a function of two variables with a function of one variable. I've already proved that it's false. What more can one say? It obviously cannot be true.

 

[redtree]

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

 

[vela]

Frankly, any function can be written in this fashion (and the same for any derivative), such that
$$ f(y) = \delta(x-y) f(x) $$

No, it can't. You can only say
$$\int_{-\infty}^\infty \delta(x-y) f(x)\,dx = f(y).$$ Note that inclusion of the integral eliminates the problem @PeroK noted. Both sides are functions of only $y$.

Moreover, sloppily speaking, one has
$$\delta(x) = \begin{cases}
+\infty & x = 0 \\
0 & x \ne 0
\end{cases}$$ 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.