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For $\alpha$ with $|\alpha|=2$ let $P$ be a homogenous harmonic Polynom of degree $2$ with $D^\alpha P\ne0$ (e.g. take $P=2x_1x_2$). Choose $\eta\in C^\infty_0(\{x:|x|<2\})$ with $\eta=1$ when $|x|<1$ and $\eta=0$ when $|x|\geq2$, set $t_k=2^k$ and $c_k=\frac1k$ with $\sum c_k$ divergent. Define $f(x)=\sum\limits_0^\infty c_k\Delta(\eta P)(t_kx)$.

How do you proof that $f$ is continuous but that $\Delta u=f$ does not have a $C^2$ solution in any neighborhood of the origin?

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For future references: Gilbarg and Trudinger, exercise 4.9(a). –  user31373 Jun 7 '12 at 17:39
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up vote 7 down vote accepted

Since $P$ is a harmonic polynomial, you have that $\triangle P = 0$. Therefore $$ \triangle(\eta P) = \triangle \eta \cdot P + 2\nabla\eta \nabla P $$ Observe that in particular $$ \operatorname{supp}(\triangle(\eta P)) \subset \{ 1\leq |x| \leq 2\}~. $$ Therefore for every $x \neq 0$, there exists a small neighborhood $V_x$ of $x$ (ball of radius $\frac12 |x|$, say), such that $f(x)$ can be expressed as a finite sum of continuous functions on $V_x$ (that is, for all but finitely many of that $\triangle(\eta P)(t_k x)$ are zero on $V_x$). So $f(x)|_{V_x}$ is a continuous function away from 0. For $x = 0$, observe that the definition requires $f(0) = 0$. It is easy to check that, since $c_k \searrow 0$, and since we have that $f(x) = c_k \triangle(\eta P)(t_k x)$ for $2^{-k} \leq |x| < 2^{-k+1}$, that $f(x) = O\left( \frac{1}{\lvert\log_2 |x|\rvert}\right)$ and hence is continuous at 0.

Now suppose $u$ solve $\triangle u = f$. Away from the origin, define the function $v$ to be $\sum \frac{c_k}{t_k^2} (\eta P)(t_k x)$. Since $|x| > 0$, only finitely many terms are non-zero in the sum (up to $k \approx - \log_2 |x|$). A similar argument as above shows that for any $|x| > 0$ there is a neighborhood of $|x|$ such that $v$ must be twice continuously differentiable, being the sum of a finite number of $C^2$ functions there.

On the other hand, since $t_k$ decreases geometrically, we have that the infinite sum expression defining $v$ is absolutely convergent. And thus $v$ is continuous.

Furthermore, as away from the origin $v$ is given by a finite sum, we can differentiate term by term in the sum. This shows that away from the origin $\triangle v = f$. Hence any continuous solution of $\triangle u = f$ must be $v$ plus some harmonic (and hence $C^\infty$) function.

It suffices to show that the Hessian of $v$ is not continuous. It is here we use the condition that $D^2 P$ does not vanish identically. Let us do it for the case of $P = x_1 x_2$. A direct computation of $\partial^2_{x_1x_2} v$ shows that away from the origin, it is given by

$$ \sum c_k \eta(t_k x) + c_k \partial_1\eta(t_k x) t_kx_1 + c_k \partial_2 \eta(t_k x) t_k x_2 + c_k \partial^2_{12}\eta(t_k x) t_k^2 x_1x_2 $$

The same argument as before shows that the last three terms are only non-zero when $t_k |x| \approx 1$. This means that their contribution to the sum is only $O(c_k) = O(\frac{1}{- \log_2|x|})$, which decays as $x\to 0$. The main first term, however, is bounded below by

$$ \sum_{k < - \log_2 |x|} c_k > \frac12 \ln\left( -\log_2 |x|\right)$$

which diverges as $x\to 0$. In particular, this means that $\partial^2_{1,2}v$ is unbounded as $x\to 0$, showing that the Hessian of $v$ cannot be continuous at the origin.

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Is this a new thing, using $\triangle $ for Laplacian? –  user53153 Feb 14 '13 at 16:25
    
@5PM: no. At least I have always used $\triangle$ for the Laplacian. It makes more sense then $\Delta$ when juxtaposed against $\Box$ for the d'Alembertian operator. (Also in relativity $\Delta$ is often used for other things, so this helps reduce confusion.) Nobody complained so far. It is slightly unfortunate that the triangle in the STIX fonts (or whatever it is that MathJax is loading) is huge. –  Willie Wong Feb 14 '13 at 16:39
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