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Let $L$ be an operator on $C^2(\mathbb R)$, defined by $$L \phi (x) = \int_{|y|<1} (\phi(x+y) - \phi(x) - \phi'(x) \ y)\ \nu(dy), \text{ for all } x\in \mathbb R$$ for a measure $\nu(dy) = |y|^{-2} dy$. It is evident that $L\phi$ is well-defined for $\phi\in C^2$ since$$|\phi(x+y) - \phi(x) - \phi'(x) \ y| \le K_x y^2$$in $B_1(x)$. Note that the definition of $L$ is explicitly dependent up to the first derivative of $\phi$.

Question. Is there a function $\phi \in C^1 \setminus C^2$, such that $L \phi $ is not well-defined?

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    $\begingroup$ It's not clear what the precise technical meaning is that you want to attach to "well defined;" can you please clarify. $\endgroup$
    – Christian Remling
    Commented Jun 5, 2016 at 16:55
  • $\begingroup$ In particular, following @ChristianRemling's question, do you want $L\varphi$ to not merely be "(well?-) defined" in some ill-defined sense, but a continuous map from $C^2(\mathbb R)$ to another topological vector space, and you're asking whether it extends to be a continuous map from $C^1(\mathbb R)$ to a perhaps-larger TVS? Or, perhaps, is the refinement to this meta-question the question you wanted to ask? $\endgroup$
    – paul garrett
    Commented Jun 5, 2016 at 21:51
  • $\begingroup$ Is it $L$ you call Laplacian?? Seems unusual... $\endgroup$
    – Jean Duchon
    Commented Jun 6, 2016 at 13:24

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$L\phi$ is a convolution with a distribution with compact support, thus it is "well defined" at least as a distribution, for any $\phi\in\cal D'$. A closer look reveals $L$ maps $H^s$ to $H^{s-1}$, $\forall s\in\mathbb R$. Indeed, this operator is a truncated version of the (positive) square root of (minus) the Laplacian, that multiplies $\hat\phi(\xi)$ with a constant times $|\xi|$. (It doesn't map $C^1$ to $C^0$, however).

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