Proof $f(x)\delta (x)=f(0) \delta(x)$

Question

How can I prove the following relation?

$$f(x)\delta (x)=f(0) \delta(x)$$

My book says that it arises from:

$$\int_{-\infty}^{+\infty}{f(x)\delta(x)dx=f(0)}$$

Thank you very much.

Answer 1

For $x = 0$, trivially $f(x)\delta(x) = f(0)\delta(x)$.

For $x \neq 0$, $\delta(x) = 0$, so $f(x)\delta(x) = 0 = f(0)\delta(x)$.

Answer 2

$\delta$ is a distribution, not a function; your first equation means that \begin{align*} \int_{-\infty}^\infty f(x)g(x)\delta(x) = \int_{-\infty}^\infty f(0)g(x)\delta(x) \end{align*} for any (suitably well-behaved) function $g$, where by definition $\int_{-\infty}^\infty \phi(x)\delta(x) = \phi(0)$. And clearly $f(x)g(x)$ and $f(0)g(x)$ agree at $x = 0$.

Answer 3

A more rigorous proof would not write the application of $\delta$ as an integral. Using the notation I learnt the proof would look like follows:

Let $X$ be an open set, $f \in C^\infty(X)$ and $\phi \in C_c^\infty(X)$. Then, $$\langle f \delta, \phi \rangle = \langle \delta, f \phi \rangle = (f\phi)(0) = f(0) \phi(0) = f(0) \langle \delta, \phi \rangle = \langle f(0) \delta, \phi \rangle$$ Since this is valid for any $\phi \in C_c^\infty(X)$ we have $f\delta= f(0)\delta$.