what does it mean that "f(x) has a delta function"? I've found some lecture on quantum mechanics and the professor is talking about functions being continuous, having jumps, or having delta functions. What does it mean to have a delta function? 
I know what Dirac delta function is, I would understand someone saying that "f(x) is a delta function" as "f(x) is a Dirac delta function". But what does it mean to say that f(x) has a delta function?
 A: What it roughly means is that in addition to discontinuities of the function which are finite jumps, there are also discontinuities of the form of dirac deltas, i.e. $"f(x)=f_1(x) + \sum_{i=1}^n\ \alpha_i \delta(x-x_i)"$, where $f$ is an e.g. piecewise continuous function and the sum is a sum of (scaled) Dirac deltas at points $x_1,  x_2 ,\dots, x_n.$ But then, this $f$ is not a function anymore.
It's rather tricky to answer this question rigorously, because the usual mathematical frameworks where you would have an object like a Dirac Delta, the set of objects considered are either so-called distributions (which means functionals on function spaces), or measures (i.e. things you integrate with respect to). Both classes of objects are generalizations of functions - so not every distribution is a function - the Dirac Delta is such an example of a non-function.
Since each measure defines an integral, and every integral is a linear functional, these notions are closely connected.
To see how discontinuities behave, let's consider a continuously differentiable function $F:\mathbb R \rightarrow \mathbb R$ and let's define an integral w.r.t. $F$ by defining for any continously differentiable (as often as you like) function $g$ with compact support:
$$\int_{\mathbb R}g(x)d(F(x))= \int_{\mathbb R}g(x)F'(x)d x.$$
(If you know some probability theory, then you'll recognize $F$ as a cumulative distribution function and $F'$ as probability density function).
But what if $F$ has a discontinuity? Consider the Heaviside function $\theta(x)=\begin{cases}0, x < 0 \\ 1, x \geq 0\end{cases}$: Then we can't use the formula above, as $\theta'$ is not well-defined. On the other hand, if we look at the right hand side, what we can do in the case of a cont. differentiable function $F$, is integration by parts and we end up with $$\int g(x)F'(x)dx = -\int g'(x)F(x)dx$$ (no boundary terms since $g$ has compact support). Now the right-hand- side is well-defined for functions $F$ which are not continuously differentable (piecewise continuous is enough), and for the choice above, we get:
$-\int g'(x) \theta(x) dx = -\int g('x) \theta(x) dx = -\int_{[0,\infty)}g'(x)dx = -\left[g(x)\right]_0^{\infty}= g(0)$
So, be defining $\int g(x)d(F(x)) = -\int g'(x) F(x) dx$, we have:
$$\int g(x) d(\theta(x)) = g(0)$$ In the sense of the definition above, the delta distibution which sends $g$ to $g(0)$ is the derivative (so.called distributional derivative) of the Heavyside function and we could write  $g(0) = \delta(g)$.
If we relax our rigor, we might write also formally (!) $"g(0) = \int g(x)\delta(x) dx"$. (Don't do this!)
Instead of the heaviside function, we might also consider functions which are either equal to $0$ or $1$ and have a finite number of jumppoints $x_1,x_2,\dots, x_n$. Then, we can do the same trick above (split the integral at $x_1,x_2,\dots, x_n$ not just at $0$) and get that the distributional derivative is the sum of dirac deltas at the points $x_1,x_2,\dots,x_n$. 
Finally, if the function is not constant between jump points, then we write $F=F_1 + \sum_{i=1}^n \alpha_i \theta(x-x_i)$, where $\alpha_i$ accounts for the jump height at $x_i$ and $F_1$ is continuous. Then, we can calculate 
$$\int g(x) d(F(x))= - \int g'(x)F_1(x) dx + \sum_{i=1}^n \alpha_i g(x_i),$$
thus the distributional derivative of $F$ is a continuous function plus some added dirac deltas - if $F_1$ is continuously differentiable.
A: Here is an example of an interesting distribution:
$$\Phi[\varphi] = \frac{1}{2} \varphi(0) + \int_{-\infty}^{\infty} \left( \frac{1}{2 \sqrt{\pi}} e^{-x^2} \right) \varphi(x) \, \mathrm{d} x $$
The distribution $\Phi$ (which, incidentally, is a probability distribution) is interesting, because at most places it is given by a density function: the smoothly varying quantity $e^{-x^2}/(2\sqrt{\pi})$. However, at the origin, $\Phi$ is given by (half of) a delta distribution.
In fact, we might make the common abuse of notation to write a "density function" for $\Phi$:
$$ \Phi(x) = \frac{1}{2 \sqrt{\pi}} e^{-x^2} + \frac{1}{2} \delta(x) $$
It would be reasonable to describe the shape of $\Phi$ has being continuous everywhere except the origin, where it has a singularity of dirac delta type.
