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Is there a function $f(x)$ defined on an open set of $\mathbb{R}^{2}$ ($ k\geq2 $) such that

1) $f$ is $C^{2}$ smooth,

2) $f$ is subharmonic, i.e. the laplacian $\Delta f$ of $f$ is positive,

3) $f$ is not real analytic?

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2 Answers 2

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Define $f(x,y)= |x|^3 + x^2.$ Then $f$ is $C^2$ and $\Delta f(x,y) = 6|x|+2.$ If $f$ were real analytic, then its Laplacian would be real analytic. But clearly this fails.

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In the whole plane, take the standard (smooth, radial, positive, supported on $B_1(0)$) mollifier $\rho$ and the fundamental solution of the Laplace operator $\Gamma$, s.t. $\Delta\Gamma=-\delta$, where $\delta$ is the Dirac delta (you can do this in all dimensions). The convolution $u=-\rho *\Gamma$ is smooth and subharmonic: $-\Delta (\rho *\Gamma)= -(\Delta \Gamma)*\rho=\delta*\rho\geq 0$. It is not analytic because $\rho *\Gamma(x)=\Gamma(x)$ outside the unitary ball, as $\Gamma$ is harmonic away from $0$ (standard regularization preserves harmonic functions) and thanks to unique analytic continuation.

You can do it more generally taking $u=-\Gamma*\phi$ for, say, a positive test function $\phi$

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