# How to formulate mathematically a piecewise function which is infinite at some specific value of its support?

Consider the following function defined for $x\ge 0$ real $$f(x) = \begin{cases} g(x) & x < x_0 \, ,\\ \infty & x= x_0 \, , \\ 0 & x > x_0 \, , \end{cases}$$ wherein $g(x)$ is a known function and $x_0 \ge 0$.

In particular, if $x_0=0$, then $f(x)$ may be represented as a kind of delta Dirac function.

I was wondering whether this piecewise function can be defined based on some known function. This will be useful for my further analysis.

Thank you.

Federiko

## 1 Answer

The object $f$ you aim to define is not a "function" $f: \mathbb{R} \to \mathbb{R}$ in the usual sense, but actually a distribution, since $\infty$ is not a real number. Also the "delta function" is not a function in the strict sense, and it is actually defined as a distribution. The latter is an object that operates on functions $\phi: \mathbb{R} \to \mathbb{R}$, and for the delta function this is usually written as

$$\delta[\phi] = \int_{-\infty}^{\infty}\delta(x)\phi(x)dx = \phi(0)\,.$$

So in your case you could define $f$ as (if this is the kind of infinity you want at $x_0$):

$$f[\phi] = \int_{0}^{\infty}f(x)\phi(x)dx = \phi(x_0) + \int_{0}^{x_0}g(x)\phi(x)dx\,.$$

• Here $x\ge 0$, why to you integrate from $-\infty$? Thanks. – Math Student Oct 21 '16 at 9:11
• Sorry, that was a typo from copy-pasting the formula for $\delta$. The integral should start at 0. – Xenos Oct 21 '16 at 9:20
• Honestly I cannot view the meaning of $\phi$. Can we chose it arbitrary? or what? Thanks! – Math Student Oct 21 '16 at 9:38
• Yes, $\phi$ is in some sense arbitrary (one usually takes it from some particular function space - smooth, compact support or the like). It is a so-called "test function" which is used to somehow "probe" the distribution. By allowing all possible test functions and specifying how the distribution $f$ (or $\delta$) behaves if it is evaluated on each of them (as in the formulas in my answer), you specify the distribution itself. – Xenos Oct 21 '16 at 15:45