Why is $f(x) \delta(x) = f(0)\delta(x)$ only true when $x=0$?

Question

This is a follow up from a previous question asked by me.

I know that $$\delta(x) = \begin{cases} 0 & \space \mathrm{for} \space x \ne 0 \\\infty&\ \mathrm{for} \space x = 0 \end{cases} $$ and that $$\int_{-\infty}^{\infty} \delta(x) \mathrm{d}x = 1$$

I also know that the product $$f(x) \delta(x)= 0\space\forall \space x\ne 0$$

I can summarize my lack of understanding with basically two questions:

$\color{red}{\mathrm{Question} \space1:}$ If I'm allowed to write $f(x) \delta(x) = f(0)\delta(x)$ then why not $f(x) \delta(x) = f(3)\delta(x)$ since $3 \ne 0$?

$\color{blue}{\mathrm{Question} \space2:}$ Also, why do we just substitute $x=0$ into the function $f(x)$ and not $\delta(x)$? In other words why don't we write $f(0)\delta(0)$ or $f(3)\delta(3)$ instead of $f(0)\delta(x)$ and $f(3)\delta(x)$ respectively. I know that $f(0)\delta(0)$ is undefined, but the point is that the $\delta$ still takes $x$ as its argument as well as $f$.

(As ever, apologies for the abuse of notation; this Dirac measure is all very new to me, hence all the questions about it)

Answer 1

PRIMER:

In This Answer and This Answer, I provided more detailed primers on the Dirac Delta. Herein, we condense the content of those answers.

The Dirac is not a function, but rather a Generalized Function also known as a Distribution.

The symbol $\int_{-\infty}^{\infty}\delta (x)f(x)\,dx$ is ,in fact, not an integral. It is a Functional that maps a test function $f$ into the number given by $f(0)$. (Note that whereas a function is a mapping, or "rule" that assigns to a number in a domain, a number in the range, a functional is a "rule" that assigns to functions in a vector space domain, a number.

We write

$$\int_{-\infty}^{\infty}\delta (x-a)f(x)\,dx=f(a)$$

but alternatively, and more compactly, we can write

$$\langle \delta_a,f \rangle=f(a)$$

For $a=0$, we have

$$\int_{-\infty}^{\infty}\delta(x)f(x)\,dx=\langle \delta_0,f \rangle=f(0)$$

Now, for this specific question, we have

$$\begin{align} \int_{\infty}^{\infty}f(x)\delta(x)\,dx&=f(0)\\\\ &=\int_{-\infty}^{\infty}f(0)\delta(x)\,dx\\\\ &\ne f(3)\\\\ &=\int_{-\infty}^{\infty}f(3)\delta(x)\,dx\end{align}$$

Therefore, $f(x)\delta(x)=f(0)\delta(x)\ne f(3)\delta(x)$ as was to be shown.

Answer 2

Note that the "definition" of $\delta(x)$ that you cite is more of an informal description, to aid with intuition. The actual definition is by the relation $$\int f(x)\delta(x)\,\mathrm dx = f(0)$$

Question 1:

We have, by definition, $$\int f(x)\delta(x)\,\mathrm dx = f(0)$$ We also obviously have $$\int f(0)\delta(x)\,\mathrm dx = f(0)$$ but $$\int f(3)\delta(x)\,\mathrm dx = f(3) \ne f(0)$$

Question 2:

Basically you already answered your question yourself: That replacement would be undefined, and thus especially not give the same value.

Answer 3

It can be true that $f(x)\delta(x) = f(0)\delta(x)$, depending on your initial definition of $\delta(x)$. But it is not the case with your definition.

Really, $\delta(x)$ is not a function. This seems to be causing you a bit of trouble. It is something termed a "distribution", which morally means it is defined only via how it acts with other functions when integrated against. The defining characteristic of $\delta(x)$ is that $$ \int_\mathbb{R} f(x) \delta(x) dx = f(0).$$

I can imagine difficulties arising from courses or books trying to use the Dirac delta distribution without really trying to understand it (since it's useful in differential equations, for instance, even if just treated as a black box).

It is true however that $$ \int_\mathbb{R} f(x) \delta(x) dx = \int_\mathbb{R} f(0) \delta(x)dx,$$ and neither of these are equal to $$ \int_\mathbb{R} f(3) \delta(x) dx.$$