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Assume function $f$ has a discontinuity (jump discontinuity) at $x_0$. This discontinuity should be buried to level $k$, meaning (let $f$ is modified to $g$ after an operation) $g$, $g'$, $g'',\ldots g^{(k-1)}$ ($g^{(i)}$ means $i$th derivative of $g$) should all be continuous at $x_0$. What is the operation on $f$ by which this an be achieved (for example convolving with a suitable function, etc.)? For a slightly different question, assume $f$ is continuous at $x_0$ but not differentiable. EDIT: $f$ has a compact support or atleast belong to $L^1$ or $L^2$. same should be the case with $g$.

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What possibilities have you considered? –  Mariano Suárez-Alvarez Nov 6 '10 at 18:53
@Rajesh D: What should be relation between the original $f$ and the new-found $f$? Do you want the $k$th derivative of your new function to equal your original one? If so, you should change the names of the new ones. –  Arturo Magidin Nov 6 '10 at 19:37
@Rajesh D: But you still have not told us the relation that you expect between $f$ and $g$. Do you want $g^{(k)}=f$? Otherwise, just pick $g=x$ for any $f$ and it will work, with the modification being "ignore $f$, just use this function instead". –  Arturo Magidin Nov 6 '10 at 19:51
Why do you want to have discontinuous derivatives, anyway? What are you really trying to do? –  J. M. Nov 7 '10 at 0:34
@Rajesh D: please put all the details about what you want in the body of the question. Explain exactly what you want to achieve and what you consider an answer. Replying to answers with "the answer is not generic enough" is really unbecoming, when you have given absolutely no description of what is "generic enough"! –  Mariano Suárez-Alvarez Nov 8 '10 at 6:42

1 Answer 1

If $f(x)$ has a jump discontinuity at $x_0$, then $g_k(x) = f(x)(x-x_0)^{k+1}$ has a continuous $k$th derivative at $x_0$ but is not $(k+1)$ times differentiable at $x_0$.

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the answer is not generic enough –  Rajesh D Nov 7 '10 at 4:04
@Rajesh: Please be more specific and say what you're looking for. –  Ryan Budney Nov 7 '10 at 4:58
Please see: math.stackexchange.com/questions/9243/… –  Rajesh D Nov 7 '10 at 5:42

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