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Is there a compactly supported and smooth (except at one point) function $f : \mathbb{R} \to \mathbb{R}$ where (non smooth point) it is is exactly k times differentiable ?

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Yes. Just take $f(x)=0$ for $x<0$, $f(x)=x^{k+1}$ for $x \ge 0$, and multiply by a smooth transition function (say one that is $1$ for $x<1$ and $0$ for $x>2$) to make it compactly supported.

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    @Chris Eagle : I am not able to get how it satisfies the condition of exactly k-times differentiability condition.Could you please explain ?2011-01-18
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    @Rajesh: I suggest taking $k$ derivatives and seeing what happens.2011-01-18
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    @Jonas Meyer : all order derivatives vanish at 0 and 1, .apperently. By k times differentiable i mean only first k derivatives exist and the higher ones do not exist.2011-01-18
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    The $(k-1)$th derivative is $0$ for $x \le 0$ and $(k+1)!x$ for $x \ge 0$. Thus the $k$th derivative fails to exist at $0$.2011-01-18
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    @Jonas Meyer : I have verified for k = 1; the first and second order derivatives are zero at 0-,0+,1-,1+. It is very simple, both functions (polynomial and bump) are smooth hence the product is also a smooth function !2011-01-18
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    @Chris Eagle : you need to take the derivative of the product function !2011-01-18
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    @Rajesh: Note that Chris has made things simple by specifying that the smooth function is the constant $1$ in a neighborhood of $0$, so there is no need to worry about the product where the differentiability is in question. I don't know what you're saying about 1- and 1+. The first $k$ derivatives are $0$ at $0$, but as Chris mentioned (with a typo) the $k^\text{th}$ derivative is not differentiable at $0$.2011-01-18
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    @Rajesh D: you were using the wrong bump function. I've changed my answer to use a transition function, which makes things a bit cleaner.2011-01-18
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    @Chris Eagle : Thank you for the clarification...i did'nt consider that you have tappered the $f(x)$ to zeros for $x<=0$. If i have understood correctly, a ordinary shifted bump function with the maximum value sitting over $x=0$ would do.(as $f(x)=0 /forall x<=0$ please confirm whether i am correct. thank you2011-01-18
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    @Rajesh D: yes, that was the sort of bump I had in mind.2011-01-18
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    @Chris Eagle,@Jonas Meyer : In this example the first (k-1) derivatives at the desired point are zero. Is this the only way of generating such a function ?2011-01-18
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    No. Let $a_0 \ldots a_k$ be the values you want for the 0th through k'th derivative. Use that to generate a Taylor polynomial of degree k and call it $g(x)$. Add this $g$ to the above $f$ before multiplying by the bump function, now you have a function with prescribed first k derivatives at the origin but whose (k+1)st derivative there does not exist.2011-01-18
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    @Willie Wong : thank you2011-01-18