The Dirac Delta Function: How do you prove the $f(0)$ property using rigorous mathematics? Dirac, Reference 1, pg 59, says
The most important property of $\delta(x)$ is exemplified by the following equation
\begin{equation*}
 \int_{-\infty}^\infty  f(x) \delta(x) dx     =f(0),~~~~~~~~~~~~~~~~~~~(3)
\end{equation*}
What you might consider, as really going on here, is that the left hand side is a notation for a limiting process. We could replace the above with
\begin{equation*}
\lim_{\epsilon\to 0^+}   \int_{-\infty}^\infty \delta_\epsilon(x) f(x) dx     =f(0)
\end{equation*}
Where the variable $\epsilon$ is used by Dirac, see pg 58 of Reference 1, if you can
Using a variable '$t$' , rather than $\epsilon$, the real meaning of (3) might be taken to be represented by
\begin{equation*}
\lim_{t\to 0^+}   \int_{-\infty}^\infty \delta_t(x) f(x) dx     =f(0)
\end{equation*}
where,
$$ \delta_t(x) = \left\{
\begin {array}{lr}
\frac{1 }{2t } & \text{if} -t \leq x \leq t     \\
0 &  \text{if    } ~~~~~~|x|>t\\
\end{array}
\right .
$$
$ \delta_t(x) $ could be called a Top-Hat function ( t for Top-Hat) because of the shape of it's graph.
Define the function $G(t)$ by
\begin{equation*}
G(t)=   \int_{-\infty}^\infty \delta_t(x) f(x) dx 
\end{equation*}
The idea of a limit of a function at a point is a "two-sided thing". For a number,'l', to be the limit of a function $G(t)$, at some point $t_0$, we have to be able to solve, for any positive $\epsilon$, however small
\begin{equation*}
|G(t)-l|<\epsilon
\end{equation*}
for't' values in some union of open intervals,
\begin{equation*}
t \in (t_0-\delta,t_0) \cup (t_0,t_0+\delta)
\end{equation*}
where in general the positively valued $\delta$ is a function of $\epsilon$.
In our case, there will exist, at best, a "one-sided limit".
What we need to show, put  $t_0=0$, is that
\begin{equation*}
\lim_{t\to 0^+}   \int_{-\infty}^\infty \delta_t(x) f(x) dx     =f(0)
\end{equation*}
or
\begin{equation*}
\lim_{t\to 0^+}G(t)=f(0)
\end{equation*}
My question is, how do we show
\begin{equation*}
\lim_{t\to 0^+}G(t)=f(0)
\end{equation*}
using rigorous mathematics.
You might answer this question for complex valued functions $f(x)$, or restrict your answer to real-valued $f(x)$.
Reference

*

*P.A.M. Dirac, The Principles Of Quantum Mechanics 4th Ed., Clarendon Press, Oxford, 1958

 A: My answer is restricted to the case of real-valued functions $f(x)$.
To prove
\begin{equation*}
\lim_{t\to 0^+}G(t)=f(0)
\end{equation*}
We need to show, that for any pre-chosen positive $\epsilon$, however small, we can find a set of values of '$t$', that satisfy the inequality
\begin{equation*}
|G(t)-f(0)|<\epsilon
\end{equation*}
these values of $t$ should form an open interval of the form $(0,\delta)$. Where in general $\delta$ is a function of $\epsilon$.
NB: f(x) is a continuous function.
Hence, for any pre-chosen positive $\epsilon$, however small, we can find what we might call a '$\tilde{\delta}$ neighborhood of $x_0$', $\tilde{\delta}_N$, at each $x_0$ in $f's$ domain
\begin{equation*}
\tilde{\delta}_N=(x_0-\tilde{\delta},  x_0+ \tilde{\delta})
\end{equation*}
For any $x\in\tilde{\delta}_N$, we have
\begin{equation*}
|f(x)-f(x_0)|<\epsilon
\end{equation*}
Put $x_0=0$.  For any allowed $\epsilon$, our chosen $\epsilon$ determines a number $\tilde{\delta}$, associated with the continuity of $f$, we then consider $G(\tilde{\delta})$
\begin{align*}
 G(\tilde{\delta})&=  \int_{-\infty}^\infty \delta_{\tilde{\delta} }(x) f(x) dx\\
&= \int_{-\tilde{\delta}   }^{\tilde{\delta}}        \delta_{\tilde{\delta} }(x) f(x) dx
\end{align*}
because,
\begin{equation*}
 \delta_{\tilde{\delta} }(x)=0~  \text{if}~ |x|>\tilde{\delta} 
\end{equation*}
The above integral, by $f's$ continuity is bounded as below ( for real-valued $f(x)$ only )
\begin{equation*}
f(0)-\epsilon \leq \int_{-\tilde{\delta}   }^{\tilde{\delta}}        \delta_{\tilde{\delta} }(x) f(x) dx\leq f(0)+\epsilon
\end{equation*}
Or, put another way,
\begin{equation*}
f(0)-\epsilon \leq  G(\tilde{\delta})  \leq f(0)+\epsilon
\end{equation*}
Hence,
\begin{equation*}
|G( \tilde{\delta}   )-f(0)|<\epsilon
\end{equation*}
Now, for $0<t<\tilde{\delta}$, we must have
\begin{equation*}
|G( t   )-f(0)|<\epsilon
\end{equation*}
i.e. We have shown, that for any positive $\epsilon$, however small, we can find a set of values of 't' that satisfy the inequality
\begin{equation*}
|G(t)-f(0)|<\epsilon
\end{equation*}
and, that this set of values of t,  has the form of an open interval, $(0, \tilde{\delta})$.
Hence,
\begin{equation*}
\lim_{t\to 0^+}G(t)=f(0)
\end{equation*}
Or, put another way,
\begin{equation*}
\lim_{t\to 0^+}   \int_{-\infty}^\infty \delta_t(x) f(x) dx     =f(0)
\end{equation*}
So, if we accept, that the above is the real meaning of
\begin{equation*}
 \int_{-\infty}^\infty \delta(x) f(x) dx     =f(0)
\end{equation*}
then, we have proved the f(0) property, for real-valued f(x), of the Dirac delta function, using rigorous mathematics.
A: We have
$$
G(t)
= \int_{-\infty}^\infty \delta_t(x) f(x) \, dx 
%= \int_{-\infty}^\infty \frac{1}{2t}\mathbf{1}_{[-t,t]}(x) f(x) \, dx 
%= \int_{-t}^{t} \frac{1}{2t} f(x) \, dx 
= \frac{1}{2t} \int_{-t}^{t} f(x) \, dx 
.
$$
Changing variable by $x=ty$ and assuming that $f$ is continuous at $x=0$ so that $f(tx) \to f(0)$ as $t\to 0$ we get
$$
G(t)
%= \frac{1}{2t} \int_{-1}^{1} f(ty) \, t \, dy
= \frac{1}{2} \int_{-1}^{1} f(ty) \, dy
\to \frac{1}{2} \int_{-1}^{1} f(0) \, dy
= f(0)
.
$$

In the rigorous theory, distributions like $\delta$ are defined as linear functionals over a space of nice "test functions" that are infinitely differentiable and vanishing quickly when $|x|\to\infty$. The action of a distribution $u$ on a test function $\phi$ is often denoted by $\langle u, \phi \rangle.$ This can be thought of as $\int_{-\infty}^{\infty} u(x) \, \phi(x) \, dx.$ Operations on distributions are then often defined by moving the operation to the test function. For example, derivation is defined by $\langle u', \phi \rangle := -\langle u, \phi' \rangle,$ based on integration by parts. In this theory, $\delta$ is defined by $\langle \delta, \phi \rangle = \phi(0).$ You can get an introduction to the theory in Lecture notes on Distributions by Hasse Carlsson.
