TITLE: Confusion with derivative of the Dirac Delta function.
QUESTION [9 upvotes]: So I was told by my instructor that $$L(\delta(t)) = 1 $$
And that $$\delta * f(t) = f(t)$$
For any $f(t)$ So $$\delta *1 = 1.$$
But this is $$\int_0^{t}\delta(z)dz.$$
So $$(1)'  = 0 = (\int_0^{t}\delta(z)dz)' = \delta(t).$$
But then wouldn't we have $$L(\delta) = 0\neq 1?$$
Am I misunderstanding something? How do we reconcile these facts?
I am using the following definition of the convolution: $$f*g = \int_0^t f(t-z)g(z)dz$$

REPLY [5 votes]: First you need to understand what the delta "function" is. Let's construct it as a limit. Let
$$f_n(x) = n \,\mathbf{1}_{[-1/2n,1/2n]}(x), $$ 
with $\mathbf{1}_A$ the indicator function of the set $A$. The functions $f_n$ are simple step functions, which take the value 0 for $|x|>1/2n$ and the value $n$ for $|x|\le1/2n$.
Now let's define the function
$$ f(x) = \lim_{n\to\infty}f_n(x)$$
and wonder what meaning can this limit have. Trivially we see that if $x\ne0$ then the pointwise limit $\lim_{n\to\infty}f_n(x)$ makes sense and is 0.
What happens when $x=0$? Then we have $\lim_{n\to\infty}f_n(0)=\infty$. This is implies that we were wrong to assume that $f(x)$ was a functions to begin with. Also the fact that the pointwise limit at 0 is infinity implies that we cannot have any form of stronger convergence. So let's try weaker ones.
First notice that for all $n$ we have
$$ \int_{-\infty}^\infty f_n(x) dx =1 $$
so we can reasonably define
$$ \int_{-\infty}^\infty f(x) dx =1. $$
So $f$ is not really a function but we are happy to assign a value to its integral.
Now take $\phi$ to be smooth and bounded on $\mathbb{R}$. Then what is the value of 
$$ I_n(\phi) := \int_{-\infty}^\infty f_n(x) \phi(x) dx? $$
Since $\phi$ is smooth it holds $\phi(x)=\phi(0)+O(x)$(Taylor's theorem). OK, let's plug it in:
$$ I_n(\phi)= \int_{-\infty}^\infty n \,\mathbf{1}_{[-1/2n,1/2n]}(x) \phi(0) dx +\int_{-\infty}^\infty n \,\mathbf{1}_{[-1/2n,1/2n]}(x) O(x) dx. $$
This implies that
$$ I_n(\phi)= \phi(0) +O(1/2n). $$
So we are happy to say that 
$$ I(\phi) := \lim_{n\to\infty}\int_{-\infty}^\infty f_n(x) \phi(x) dx = \phi(0). $$
Now the question is which object is the delta function, $f$ or $I$? If you are a physicist then your answer is probably $f$. If you are a mathematician then your answer is probably $I$.
The problem of the physicist's answer is that you would say things like "the delta function is infinite at 0 and 0 everywhere else", which is kind of OK but it can be problematic. Let's see that. Let
$$g_n(x) = n \,\mathbf{1}_{[-1/n,1/n]}(x) $$
and as it was done above define
$$ g(x) = \lim_{n\to\infty} g_n(x). $$
Then for the "function" $g$ we can say exactly the same thing as above. Is it however the delta function? It is not because its integral is 2 and not 1. The limit is actually 2 times the delta function.
So let's go to your question. What is the value of $ 1*\delta$? The answer depends on your definition of convolution. We saw that is reasonable to write
$$ \int_{-\infty}^\infty \delta(x)dx=1. $$
On the other hand the integral
$$ \int_{-x_0}^x \delta(t)dt $$
with some $x_0>0$ is a function of $x$. For $x<0$ is 0 and for $x>0$ is 1. We may assign a value at $x=0$ if we need. This is the Heaviside step function and it is reasonable to say that it is the antiderivative of the delta function. I took $x_0>0$ to avoid the issue at 0. We can however do the same with your definition of convolution.
A bit about hyperfunctions.
Hyperfunctions bring complex analysis in real analysis. Which I personally find extremely cool. So the idea is to represent generalised functions as the difference of 2 analytic functions.
From now on I will say things like "$f$ is analytic on $\mathbb{R}$". This means that there exists an open set $A\in\mathbb{C}$ such that it contains $\mathbb{R}$ and $f$ is analytic in $A$.
So how does it work? Obviously if you take 2 functions analytic on $\mathbb{R}$ then their difference is also analytic on $\mathbb{R}$. So we gain nothing. But on the other hand, if we take a function $F_+$ analytic on the open upper half plane and a function $F_-$ analytic on the lower half plane then their difference makes no sense since there is no domain where they are both defined. :)
However! We may be able to define the difference of their limits!
So let's start by defining what a hyperfunction is. We denote a hyperfunction by $F=[F_+,F_-]$ and we say that the hyperfunction $F$ has $F_+$ as upper component and $F_-$ as lower component, with $F_+$ and $F_-$ as above.
So how does it act on test functions? Let $\phi$ be a function analytic on $\mathbb{R}$ which decreases exponentially toward the real infinity. Let $A(\mathbb{R})$ be the space of all such functions. We see that this space is a subspace of the Schwartz space. Then we define
$$ F[\phi] = \int_{\mathbb{R}+\epsilon i} F_+(z)\phi(z)dz - \int_{\mathbb{R}-\epsilon i} F_-(z)\phi(z)dz. $$
So we integrate just above and just below the real line and we take the difference. This means that $F$ is in the dual of $A(\mathbb{R})$, so let's see how many elements of the dual we can represent.
Let 
$$I(\phi) = \int_{\mathbb{R}} \phi(z) dz $$
Then obviously we can write
$$ I = [1,0]=[1/2,-1/2]=[0,-1]. $$
This brings up an important point about hyperfunctions: let $\psi\in C^\omega(\mathbb{C})$, then
$$ F=[F_+,F_-] =[F_++\psi,F_-+\psi]. $$ 
This come from the fact that we subtract the 2 integrals. So the proper definition for the space of hyperfunctions requires a to take the modulous with a proper equivalence relation (which you can probably guess) but I will not go into details on that.
So what more can we do? Let's define
$$ J(\phi) = \int_0^\infty \phi(z) dz. $$
Trivially it holds 
$$ J(\phi)=I(H\cdot\phi), $$
with $H$ the Heaviside step function. Let's define
$$ J = [-\frac{1}{2 \pi i}\log(-z),-\frac{1}{2 \pi i}\log(-z)], $$
then I claim that $J$ is actually the Heaviside step function. I will not go througth this, but it is relatively simple to see it using the definition $\log(z) = \int_1^z \frac{1}{t}dt$. What makes this work is that for $x<0$ the value of the logarithm is the same, however for $x>0$ it depends on the path of integration.
Now let's do the delta function. We define as usual
$$\delta(\phi) = \phi(0).$$
Lets's recall Cauchy's integral that states that if a function is analytic in a neigbourhood of the origin then
$$ \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{\zeta} d\zeta = f(0) $$
and define 
$$ \delta = [-\frac{1}{2\pi i z},-\frac{1}{2\pi i z}]. $$
So how does this make any sense? First notice that for all $x\ne0$, the values of the "two" functions are the same, so the value of the hyperfunction is 0. At $x=0$ it complex infinity. But actually we can say much more things about the origin. Both functions have a simple pole with coefficient $\frac{1}{2\pi i}$. So when we integrate and take the difference, we can transform the integral to the Cauchy integral (I'm not going to actually do this here, but it is pretty straightforward).
What is cool about the theory is that we can define the derivative in the most straightforward way possible: just by writing
$$ F'=[F_+',F_-']. $$
The beauty of this can be shown by writting
$$ J' = [(-\frac{1}{2 \pi i}\log(-z))',(-\frac{1}{2 \pi i}\log(-z))'] = [-\frac{1}{2\pi i z},-\frac{1}{2\pi i z}] = \delta. $$
So the delta function is the actual derivative of the Heaviside step function!
I will stop here. There is much more into this theory and it is not an easy subject but it is a beautiful one. A relatively simple introduction is the book "Introduction to Hyperfunctions and Their Integral Transforms" by Urs Graf.