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Let $u$ be a real valued function with domain $\mathbb{R}^n$ and I want to find all pair $(p,X)\in \mathbb{R}^n\times\mathcal{S}^n$, where $\mathcal{S}^n$ is the set of symmetric $n\times n$ matrices, such that $$\lim\sup_{y\to x}\frac{u(y)-u(x)-\langle p,(y-x)\rangle-\frac{1}{2}\langle X(y-x),(y-x)\rangle}{|y-x|^2}\le 0$$ Where $\langle\cdot,\cdot\rangle$ is the canonical scalar product. Furthermore, with $|\cdot|$ we denote the euclidean norm on $\mathbb{R}^n$. Consider a concrete example. Let $u(x):=-|x|$, here $n=1$, so both $p,X$ are real numbers. Of course I have to use a case-by-case analysis. Suppose $x>0$. I have to find all pairs $(p,X)$ such that $\lim\sup_{y\to x}\frac{u(y)-u(x)-p(y-x)-\frac{1}{2}X(y-x)(y-x)}{|y-x|^2}=\lim\sup_{y\to x}\frac{-|y|+|x|-p(y-x)-\frac{1}{2}X(y-x)(y-x)}{|y-x|^2}=\lim\sup_{y\to x}\frac{-|y|+x-p(y-x)-\frac{1}{2}X(y-x)(y-x)}{|y-x|^2}\le 0$

Now how can I find all this pairs $(p,X)$? I have no idea how to solve this. Thank you for your help

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    Where have you used the function $f$? Or am I missing something?2012-07-10
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    Sorry, that was a mistake. $f$ should be $u$. I edited my question2012-07-10
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    Say $n=2$. What are $x$ and $y$? scalars, or vectors? ditto for $u(x)$. In the numerator, it looks like you are subtracting a matrix from a vector, or else multiplying two vectors.2012-07-10
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    Sorry if my notation was not clear. I added some details.2012-07-10
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    That's better, but you still have the incomprehensible $p(y-x)$, a product of some form of two vectors. What is it?2012-07-10
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    I'm really sorry. Of course this should be a scalar product too. In the example everything is just an multiplication.2012-07-10

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