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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.

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    $\begingroup$ 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. $\qquad$ $\endgroup$ – Michael Hardy Feb 18 '16 at 16:18
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    $\begingroup$ 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. $\qquad$ $\endgroup$ – Michael Hardy Feb 18 '16 at 16:25
  • $\begingroup$ @Michael Hardy: advice accepted..."any"--"every"...thanks $\endgroup$ – Sudeepan Datta Feb 18 '16 at 16:44
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To start with, part of the trouble is that a discontinuous function can determine the same distribution as a smooth function.

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.

There is also the possibility of having functions that are not almost everywhere equal to something smooth but whose discontinuities are not jumps, i.e. the left and right limits do not exist.

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