8
$\begingroup$

Given $u \in \mathcal{C}^\infty_0(\mathbb{R}^n)$, $u \geq 0$ everywhere, is $v(x) = \sqrt{u(x)}$ also in $\mathcal{C}^\infty_0$? It is clear that the only problematic points are the boundary of the support, where one must show that all the derivatives vanish.

I would appreciate any help!

  • 2
    Do you mean to ask whether there exists $v\in\mathcal{C}^\infty_0$ such that $v^2 = u$? (As the question is stated, it sounds like you're taking $v(x)$ to be the unique nonnegative square root of $u(x)$, in which case taking $n=1$ and $u$ to be $x^2$ (times a bump) is a counterexample, as $\sqrt{u}$ is locally $|x|$.2012-08-25
  • 1
    Asked and answered at http://mathoverflow.net/questions/105438/square-root-of-a-positive-c-infty-function2012-08-25
  • 0
    Yes, you are right ... That was the intent of my question, taking the unique non-negative square root. I got blind by the application I suppose. Thanks for the counterexample!2012-08-25

2 Answers 2

8

There is an example of a nonnegative $u\in\mathcal{C}^\infty_0(\mathbf{R}^2)$ which does not admit a differentiable square root. Namely, $u = (x^2 + y^2)\varphi$, where $\varphi$ is a bump localised at $0$. The only continuous square roots of $u$ are $\pm\sqrt{x^2 + y^2}\sqrt{\varphi}$, neither of which are differentiable at $0$.

  • 2
    Very good. This pretty much nails the interior-point issue.:)2012-08-25
2

Edit: my previous incorrect answer was: .... "to show that the $n$-th derivative of the square root exists (and vanishes) at a boundary point, use the existence and vanishing of all derivatives up to order $2n$ of the original function, in a Taylor-Maclaurin expansion with remainder."

Edit: But there are already problems at interior points, as comments and examples show.

  • 0
    Thanks! Can you please clarify: around which point must I take the Taylor series? The Taylor series of $\sqrt{y}$ does not exist at $y=0$.2012-08-24
  • 1
    Indeed, $\sqrt{y}$ is not smooth at $y=0$, ... but, also, $y$ does not vanish to large order there. A function vanishing to order $2n$ at a point is $n$-fold differentiable. That is, look at the Taylor expansion of the given function to estimate the square root, _showing_ that the square root is continuous, differentiable, etc.2012-08-24
  • 5
    [This MO thread](http://mathoverflow.net/questions/105438/) contains a discussion. In section 2 [here](http://www.math.polytechnique.fr/~bony/BBCP_jfa.pdf) (from Palais's answer) the function $$f(x) = \exp\left(-\frac{1}{|x|}\right) \left[\sin^2\left(\frac{\pi}{|x|}\right) + \exp\left(-\frac{1}{x^2}\right)\right]$$ is given as an example of a nonnegative smooth function vanishing only at $0$, admitting a $C^1$ square root, but no square root of regularity $C^{1,\alpha}$ with $0 \lt \alpha \lt 1$. Multiply $f$ by a smooth bump function centered at $0$ to get an example with compact support.2012-08-25
  • 1
    The example above is a variation of a construction from [this 1963 paper](http://www.numdam.org/numdam-bin/fitem?id=AIF_1963__13_2_203_0) by Glaeser.2012-08-25
  • 1
    I am shocked! :)2012-08-25
  • 0
    This is great stuff.2012-08-25