Product of dirac delta function without integral

Question

I am learning Distribution theory. I see that $\int_{-\infty}^{\infty} f(x) \delta(x) \,dx = f(0)$

The question is, at x=0, is $f(x) \delta(x) = f(0)$, even without integration? Because at one place, it says $\delta(x)$ is infinitely large at $x = 0$ ( In that case the above expression becomes $f(x)$ times $\infty$). And at other place I see expressions like $f(x) \delta(x)$ used i.e without integrals.

Answer 1

No, $f(x)\,\delta(x)$ does not equal $f(0)$ without integration. It does however equal $f(0)\,\delta(x).$

Answer 2

If one writes $f(x) \delta(x)$, it means you are seeing the outcome as a function over the real numbers. In this context, the rigoruos way of understanding this is not using the Dirac $\delta$ but the function whose value is $0$ at $1$ and $0$ otherwise. If you write $f \delta$ in distribution theory it means that we are considering the application of the distribution $\delta$ to $f$, i.e. $\langle \delta , \phi \rangle$.

If used within an integral, it is a deterministic probability measure ($\delta(x) dx = d \delta$ and the integral over the whole domain is $1$) where the outcome is $0$ with probability $1$. It happens that this later point of view is coherent with the definition of a distribution as it defines a continuous linear form over the space of tests functions by doing : $\langle \delta , \phi \rangle = \int_\mathbb{R} \phi(x) \delta(x) dx = \int_\mathbb{R} \phi(x) d \delta = \mathbb{E}(\phi(X))= \phi(0)$ with $\mathbb{E}$ the expectation operator.

PS : note that is is coherent with the other answer as $f(x) \delta(x) = f(0) \delta(x)$ no matter if you are considering the Dirac $\delta$ or the function that is $1$ only at $0$.