Apparently, the Schwartz kernel theorem states that all linear operators can be represented as integral transforms (but only if you use generalized functions such as the dirac delta as kernels.) Representing the derivative operator as an integral transform would be really useful because then when defining strange and new generalizations of integrals one could automatically derive an appropriate generalization of derivatives to fit.
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Yes, it's the derivative of the Dirac delta, isn't it? https://en.wikipedia.org/wiki/Dirac_delta_function#Distributional_derivatives – Jun 06 '16 at 20:32
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@Rahul You seem to be right https://en.wikipedia.org/wiki/Unit_doublet – Molly Stewart-Gallus Jun 06 '16 at 20:35
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You may be interested in the Riesz transform. – Jun 06 '16 at 20:39
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Yes, you can - but (somewhat disappointingly) the integral kernel for the derivative is just
$$ k(x,y) = -\delta^\prime(x-y) $$ where $\delta^\prime$ is the derivative of the delta distribution, defined in the distributional sense i.e.
$$ \langle \delta^\prime(x-y),\varphi(y)\rangle = -\langle \delta(x-y),\varphi^\prime(y)\rangle = -\varphi^\prime(x) $$
This is occasionally useful for manipulating expressions symbolically. Also look into the single and double layer potential which express a similar idea for directional derivatives across surfaces and is a very useful concept in PDE.
icurays1
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1An additional information (that @Steven Stewart-Gallus has implicitely mentionned by citing a - disappointing - Wikipedia article) is that there is nice physical interpretation of $\delta'$ as modelizing a "doublet" (two opposite charges $q^+q^-$ that are infinitely close. – Jean Marie Jun 06 '16 at 21:55
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The kernel of the nth derivative is the nth derivative of the dirac delta right? And the dirac delta is the identity kernel and the heaviside step function and its integrals are the integrals right? – Molly Stewart-Gallus Jun 07 '16 at 01:38
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