# Generalised derivatives of discontinuous real functions

Does the generalised derivative of every discontinuous real-valued function always yield a Dirac-delta 'function' at the point(s) of discontinuity? My limited experience with generalised distributions answers in the positive but I lack a general answer to this question. Please, if possible, provide me with an intuitively clear answer to this question, because I do not know the formal language used in discussing such distributions. If not possible, then provide me with a formal explanation all the same...at least something is better than none...

If the answer shall be in the negative ( which seems to be the case here), then I would like to know under what precise conditions (such as the type of discontinuity), a Dirac -delta function can occur on taking the generalised derivative.

• Here's a pet peeve: English-speaking mathematicians use the word "any" to much. In normal English, it is reasonable to construe the question "Does the generalised derivative of any discontinuous real-valued function always yield this-or-that?" to mean "Is there any discontinuous real-valued function whose generalised derivative always yields this-or-that?" But I don't think that is what was meant. Just changing "any" to "every" would fully disambiguate the sentence. $\qquad$ – Michael Hardy Feb 18 '16 at 16:18
• When the discontinuity is a jump, then the derivative will be the sum of a corresponding delta function and whatever else is involved. But not all discontinuities are jumps. Consider the function $x\mapsto\left. \begin{cases} \sin\frac1x &\text{if }x\ne0, \\ 0 & \text{if }x=0.\end{cases}\right\}$. That is discontinuous at $x=0$, but the discontinuity is not a jump. $\qquad$ – Michael Hardy Feb 18 '16 at 16:25
• @Michael Hardy: advice accepted..."any"--"every"...thanks – Sudeepan Datta Feb 18 '16 at 16:44

Consider the Dirichlet function $\chi_{\mathbb{Q}}$ (1 on rationals, 0 on irrationals), which is discontinuous everywhere. For any test function $\phi$, we have
$$\int_{-\infty}^\infty \chi_{\mathbb{Q}} \phi \, dx = 0,$$ because the rationals are only a countable set and have measure 0. Therefore from a distributional perspective $\frac{d}{dx} \chi_{\mathbb{Q}} = 0$. As you can see, this does not yield a Dirac delta at the point of discontinuity. Similarly, you can look at Thomae's function which is also distributionally zero, but has points of discontinuity at the rational numbers only.