# Square root of compactly supported C-infinity function

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!

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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 $|x|$. –  Sean Eberhard Aug 25 '12 at 11:02
Asked and answered at mathoverflow.net/questions/105438/… –  Nate Eldredge Aug 25 '12 at 17:25
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! –  Nemis L. Aug 25 '12 at 19:27

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 $0$.

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Very good. This pretty much nails the interior-point issue.:) –  paul garrett Aug 25 '12 at 13:48

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.

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Thanks! Can you please clarify: around which point must I take the Taylor series? The Taylor series of $\sqrt{y}$ does not exist at $y=0$. –  Nemis L. Aug 24 '12 at 20:16
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. –  paul garrett Aug 24 '12 at 21:06
This MO thread contains a discussion. In section 2 here (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. –  t.b. Aug 25 '12 at 10:50
The example above is a variation of a construction from this 1963 paper by Glaeser. –  t.b. Aug 25 '12 at 10:55
I am shocked! :) –  paul garrett Aug 25 '12 at 13:37